Principle of Calorimetry and Water Equivalent Questions in English

(D) Heat required to raise $1\, kg$ of ice from $-10\,^{\circ}C$ to $0\,^{\circ}C$: $Q_1 = m_i c_i \Delta T = 1 \times 2100 \times 10 = 21,000\, J$.
Heat required to melt $1\, kg$ of ice at $0\,^{\circ}C$: $Q_2 = m_i L_f = 1 \times 3.36 \times 10^5 = 336,000\, J$.
Total heat required to convert ice to water at $0\,^{\circ}C$: $Q_{total} = 21,000 + 336,000 = 357,000\, J$.
Heat available from $4.4\, kg$ of water cooling from $30\,^{\circ}C$ to $0\,^{\circ}C$: $Q_{avail} = m_w c_w \Delta T = 4.4 \times 4200 \times 30 = 554,400\, J$.
Since $Q_{avail} > Q_{total}$,the ice melts completely and the final temperature $\theta$ is above $0\,^{\circ}C$.
Heat balance equation: $Q_{avail} - Q_{total} = (m_i + m_w) c_w (\theta - 0)$.
$554,400 - 357,000 = (1 + 4.4) \times 4200 \times \theta$.
$197,400 = 5.4 \times 4200 \times \theta$.
$197,400 = 22,680 \times \theta$.
$\theta = \frac{197,400}{22,680} \approx 8.7\,^{\circ}C$.

(B) According to the principle of calorimetry,heat lost by the water is equal to the heat gained by the solid ball.
$m_w \cdot S_w \cdot \Delta T_w = m_s \cdot S_s \cdot \Delta T_s$
Given:
$m_w = m_s = 200 \, g$
$S_w = 1 \, cal/g \cdot ^oC$
Temperature change for water: $\Delta T_w = 80 \, ^oC - 60 \, ^oC = 20 \, ^oC$
Temperature change for solid: $\Delta T_s = 60 \, ^oC - 20 \, ^oC = 40 \, ^oC$
Substituting the values:
$200 \cdot S_w \cdot 20 = 200 \cdot S_s \cdot 40$
$S_w \cdot 20 = S_s \cdot 40$
$S_s = S_w \cdot (20 / 40)$
$S_s = 0.5 \cdot S_w$
Therefore,the specific heat of the solid is one half that of water.

(A) Let $m$ be the mass of ice that melts to reach equilibrium at $0\,^{\circ}C$.
Heat lost by water $= m_w \cdot c_w \cdot \Delta T = 100 \times 1 \times (30 - 0) = 3000\, cal$.
Heat gained by ice to melt $= m \cdot L_f = m \times 80$.
Equating heat lost and heat gained: $3000 = 80m$.
Solving for $m$: $m = \frac{3000}{80} = 37.5\, g$.
This is the amount of ice that melts.
The initial mass of ice was $50\, g$.
Therefore,the amount of ice left $= 50 - 37.5 = 12.5\, g$.

(C) According to the principle of calorimetry,heat lost by steam = heat gained by ice.
Heat lost by $M$ grams of steam at $100^{\circ} C$ to become water at $40^{\circ} C$:
$Q_{lost} = M \times L_v + M \times c_w \times \Delta T$
$Q_{lost} = M \times 540 + M \times 1 \times (100 - 40) = 540M + 60M = 600M$
Heat gained by $200 \; g$ of ice at $0^{\circ} C$ to become water at $40^{\circ} C$:
$Q_{gained} = m_{ice} \times L_f + m_{ice} \times c_w \times \Delta T$
$Q_{gained} = 200 \times 80 + 200 \times 1 \times (40 - 0) = 16000 + 8000 = 24000 \; cal$
Equating the two:
$600M = 24000$
$M = 24000 / 600 = 40 \; g$.

(D) Let the temperatures of water in containers $C_{1}, C_{2}$,and $C_{3}$ be $T_{1}, T_{2}$,and $T_{3}$ respectively.
Using the principle of calorimetry $(m_{1}T_{1} + m_{2}T_{2} = (m_{1}+m_{2})T_{mix})$:
$1$. For the first mixture: $1T_{1} + 2T_{2} = (1+2)60 = 180$ ---$(i)$
$2$. For the second mixture: $1T_{2} + 2T_{3} = (1+2)30 = 90$ ---(ii)
$3$. For the third mixture: $2T_{1} + 1T_{3} = (2+1)60 = 180$ ---(iii)
Adding equations $(i)$,(ii),and (iii):
$(1+2)T_{1} + (2+1)T_{2} + (2+1)T_{3} = 180 + 90 + 180$
$3(T_{1} + T_{2} + T_{3}) = 450$
$T_{1} + T_{2} + T_{3} = 150^{\circ} C$
For the final mixture of $1 \ l$ each:
$1T_{1} + 1T_{2} + 1T_{3} = (1+1+1)\theta$
$150 = 3\theta$
$\theta = 50^{\circ} C$

(C) According to the principle of calorimetry,at steady state: Heat lost by aluminium sphere = Heat gained by water + Heat gained by calorimeter.
Mass of aluminium sphere $(m_1) = 0.047 \; kg$.
Initial temperature of aluminium sphere = $100 \; ^{\circ}C$.
Final temperature = $23 \; ^{\circ}C$.
Change in temperature $(\Delta T_1) = 100 \; ^{\circ}C - 23 \; ^{\circ}C = 77 \; ^{\circ}C$.
Heat lost by aluminium sphere $Q_{lost} = m_1 s_{Al} \Delta T_1 = 0.047 \times s_{Al} \times 77$.
Mass of water $(m_2) = 0.25 \; kg$.
Mass of copper calorimeter $(m_3) = 0.14 \; kg$.
Initial temperature of water and calorimeter = $20 \; ^{\circ}C$.
Final temperature = $23 \; ^{\circ}C$.
Change in temperature $(\Delta T_2) = 23 \; ^{\circ}C - 20 \; ^{\circ}C = 3 \; ^{\circ}C$.
Specific heat of water $(s_w) = 4.18 \times 10^3 \; J \; kg^{-1} K^{-1}$.
Specific heat of copper $(s_{cu}) = 0.386 \times 10^3 \; J \; kg^{-1} K^{-1}$.
Heat gained by water and calorimeter $Q_{gained} = (m_2 s_w + m_3 s_{cu}) \Delta T_2$.
$Q_{gained} = (0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3) \times 3$.
$Q_{gained} = (1045 + 54.04) \times 3 = 1099.04 \times 3 = 3297.12 \; J$.
Equating heat lost and gained:
$0.047 \times s_{Al} \times 77 = 3297.12$.
$3.619 \times s_{Al} = 3297.12$.
$s_{Al} = 3297.12 / 3.619 \approx 911.06 \; J \; kg^{-1} K^{-1}$.
Converting to $kJ \; kg^{-1} K^{-1}$,$s_{Al} \approx 0.911 \; kJ \; kg^{-1} K^{-1}$.

(D) Heat lost by water $= m_w s_w (\theta_i - \theta_f) = (0.30 \; kg) (4186 \; J \; kg^{-1} \; K^{-1}) (50.0^{\circ} C - 6.7^{\circ} C) = 54376.14 \; J$.
Heat gained by ice to melt $= m_i L_f = (0.15 \; kg) L_f$.
Heat gained by melted ice to reach $6.7^{\circ} C = m_i s_w (\theta_f - 0^{\circ} C) = (0.15 \; kg) (4186 \; J \; kg^{-1} \; K^{-1}) (6.7^{\circ} C) = 4206.93 \; J$.
By the principle of calorimetry,Heat lost = Heat gained:
$54376.14 \; J = (0.15 \; kg) L_f + 4206.93 \; J$.
$(0.15 \; kg) L_f = 54376.14 \; J - 4206.93 \; J = 50169.21 \; J$.
$L_f = \frac{50169.21 \; J}{0.15 \; kg} = 3.3446 \times 10^5 \; J \; kg^{-1} \approx 3.34 \times 10^5 \; J \; kg^{-1}$.

(0.433 J/GK) Mass of the metal,$m = 0.20 \; kg = 200 \; g$.
Initial temperature of the metal,$T_{1} = 150 \; ^{\circ}C$.
Final temperature of the metal,$T_{2} = 40 \; ^{\circ}C$.
Water equivalent of the calorimeter,$m' = 0.025 \; kg = 25 \; g$.
Volume of water,$V = 150 \; cm^{3}$.
Mass of water,$M = 150 \; g$ (since density is $1 \; g/cm^{3}$).
Fall in temperature of the metal,$\Delta T = 150 - 40 = 110 \; ^{\circ}C$.
Rise in temperature of the water and calorimeter,$\Delta T' = 40 - 27 = 13 \; ^{\circ}C$.
Specific heat of water,$C_{w} = 4.186 \; J \cdot g^{-1} \cdot K^{-1}$.
Heat lost by metal = Heat gained by water and calorimeter.
$m \cdot C \cdot \Delta T = (M + m') \cdot C_{w} \cdot \Delta T'$.
$200 \cdot C \cdot 110 = (150 + 25) \cdot 4.186 \cdot 13$.
$22000 \cdot C = 175 \cdot 4.186 \cdot 13 = 9523.15$.
$C = 9523.15 / 22000 \approx 0.433 \; J \cdot g^{-1} \cdot K^{-1}$.
If heat is lost to the surroundings,the measured heat gained by the water/calorimeter is less than the actual heat lost by the metal. Thus,the calculated specific heat $C$ would be smaller than the actual value.

(N/A) Calorimetry is the branch of science that deals with the measurement of heat.
$A$ calorimeter is a device used to measure the amount of heat involved in a physical or chemical process.
Principle: The principle of calorimetry is based on the law of conservation of energy. In an isolated system (where no heat exchange occurs with the surroundings), the heat lost by a hot body is equal to the heat gained by a cold body. Mathematically, $\text{Heat lost} = \text{Heat gained}$.
Construction: $A$ calorimeter typically consists of a metallic vessel and a stirrer made of the same material, such as copper or aluminium. The vessel is placed inside a wooden jacket or box, which is filled with heat-insulating materials like glass wool or cotton to minimize heat loss to the surroundings. The outer jacket acts as a thermal shield. $A$ lid is provided with an opening to insert a mercury thermometer to measure the temperature changes during the experiment.

(C) Mass of water $m = 100 \, g$.
When supercooled water at $-10 \, ^\circ C$ is disturbed,it releases heat as it warms up to $0 \, ^\circ C$. This heat is used to freeze a portion of the water into ice.
Heat released by $100 \, g$ of water to reach $0 \, ^\circ C$:
$Q = m \cdot S_W \cdot \Delta T = 100 \times 1 \times (0 - (-10)) = 1000 \, cal$.
Let $m'$ be the mass of water that freezes into ice at $0 \, ^\circ C$. The heat released by the water is absorbed by the freezing process:
$Q = m' \cdot L_{fusion}$.
$1000 = m' \times 80$.
$m' = \frac{1000}{80} = 12.5 \, g$.
Since the final state is a mixture of ice and water at equilibrium,the temperature of the resultant mixture will be $0 \, ^\circ C$ and the mass of ice formed is $12.5 \, g$.

(B) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.
Heat gained by the calorimeter and water:
$Q_{\text{gained}} = (m_{\text{cal}} + m_{\text{water}}) \times c_w \times \Delta T$
$Q_{\text{gained}} = (20 \, g + 180 \, g) \times 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1} \times (31^{\circ} C - 25^{\circ} C)$
$Q_{\text{gained}} = 200 \times 6 = 1200 \, \text{cal}$.
Heat lost by steam at $100^{\circ} C$ to become water at $31^{\circ} C$:
$Q_{\text{lost}} = m \times L_v + m \times c_w \times \Delta T$
$Q_{\text{lost}} = m \times 540 + m \times 1 \times (100^{\circ} C - 31^{\circ} C)$
$Q_{\text{lost}} = m \times (540 + 69) = 609m$.
Equating heat gained and lost:
$1200 = 609m$
$m = \frac{1200}{609} \approx 1.97 \, g$.
Thus,the value of '$m$' is close to $2 \, g$.

(A) The heat lost by the water as it cools from $25^{\circ} C$ to $0^{\circ} C$ is used to melt the ice.
Heat lost by water: $Q = m_w s_w \Delta \theta$
$Q = 0.2 \, kg \times 4200 \, J \, kg^{-1} \, K^{-1} \times (25 - 0) \, K = 21000 \, J$
Heat required to melt $m_{ice}$ grams of ice: $Q = m_{ice} L_f$
$21000 \, J = m_{ice} \times 3.4 \times 10^{5} \, J \, kg^{-1}$
$m_{ice} = \frac{21000}{3.4 \times 10^{5}} \, kg$
$m_{ice} = 0.06176 \, kg = 61.76 \, g$
Rounding to the nearest value,the amount of ice that melts is approximately $61.7 \, g$.

(C) Let the specific heats of liquids $x$,$y$,and $z$ be $s_1$,$s_2$,and $s_3$ respectively. Since the masses are equal $(m_1 = m_2 = m_3 = m)$,the principle of calorimetry states: $m s_1 T_1 + m s_2 T_2 = (m s_1 + m s_2) T_f$.
For $x$ and $y$ mixed: $s_1(10) + s_2(20) = (s_1 + s_2)(16) \implies 10s_1 + 20s_2 = 16s_1 + 16s_2 \implies 4s_2 = 6s_1 \implies s_1 = \frac{2}{3}s_2$.
For $y$ and $z$ mixed: $s_2(20) + s_3(30) = (s_2 + s_3)(26) \implies 20s_2 + 30s_3 = 26s_2 + 26s_3 \implies 4s_3 = 6s_2 \implies s_3 = \frac{3}{2}s_2$.
For $x$ and $z$ mixed: $s_1(10) + s_3(30) = (s_1 + s_3)T_f$.
Substituting $s_1 = \frac{2}{3}s_2$ and $s_3 = \frac{3}{2}s_2$:
$(\frac{2}{3}s_2)(10) + (\frac{3}{2}s_2)(30) = (\frac{2}{3}s_2 + \frac{3}{2}s_2)T_f$.
$\frac{20}{3}s_2 + 45s_2 = (\frac{4+9}{6})s_2 T_f \implies \frac{20+135}{3} = \frac{13}{6}T_f$.
$\frac{155}{3} = \frac{13}{6}T_f \implies T_f = \frac{155 \times 6}{3 \times 13} = \frac{310}{13} \approx 23.84^{\circ}C$.

(C) The heat lost by the copper block as it cools from $500^{\circ} C$ to $0^{\circ} C$ is given by the formula $\Delta Q = mc\Delta T$.
Mass of copper $m = 5.0 \, kg = 5000 \, g$.
Specific heat of copper $c = 0.39 \, J g^{-1} {}^{\circ} C^{-1}$.
Temperature change $\Delta T = 500^{\circ} C - 0^{\circ} C = 500^{\circ} C$.
Heat released $\Delta Q_1 = 5000 \times 0.39 \times 500 = 975,000 \, J$.
Let $m_{ice}$ be the mass of ice that melts. The heat absorbed by the ice to melt at $0^{\circ} C$ is $\Delta Q_2 = m_{ice} \times L_f$,where $L_f = 335 \, J g^{-1}$.
Equating the heat lost and heat gained: $\Delta Q_1 = \Delta Q_2$.
$975,000 = m_{ice} \times 335$.
$m_{ice} = \frac{975,000}{335} \approx 2910.45 \, g$.
Converting to $kg$,$m_{ice} \approx 2.91 \, kg$.

(A) The heat lost by the water as it cools from $25^{\circ}C$ to $0^{\circ}C$ is given by $Q = mc\Delta T$.
Here,$m = 0.3\,kg$,$c = 4200\,J\,kg^{-1}K^{-1}$,and $\Delta T = 25^{\circ}C$.
$Q = 0.3 \times 4200 \times 25 = 31500\,J$.
This heat is used to melt $x\,g$ (or $m\,kg$) of ice at $0^{\circ}C$.
The heat required to melt ice is $Q = mL_f$,where $L_f = 3.5 \times 10^{5}\,J\,kg^{-1}$.
$31500 = m \times 3.5 \times 10^{5}$.
$m = \frac{31500}{3.5 \times 10^{5}} = \frac{31500}{350000} = 0.09\,kg$.
Converting to grams,$m = 0.09 \times 1000 = 90\,g$.
Therefore,$x = 90$.

(B) Let $m$ be the mass of steam (in grams) condensed to heat the mixture from $25^{\circ} C$ to $70^{\circ} C$.
Heat lost by steam = Heat gained by mixture.
Heat lost by steam = (Heat of condensation of steam) + (Heat released by condensed water cooling from $100^{\circ} C$ to $70^{\circ} C$).
Heat gained by mixture = (Mass of mixture) $\times$ (Specific heat of water) $\times$ (Change in temperature).
$m L + m s_w (100 - 70) = M s_w (70 - 25)$.
Given $L/s_w = 540^{\circ} C$,$M = 500 \, g$,and $\Delta T = 45^{\circ} C$.
$m(540 + 30) = 500 \times 45$.
$m = (500 \times 45) / 570 \approx 39.47 \, g$.
Rate of steam supply = $50 \, g/min = 50/60 \, g/s = 5/6 \, g/s$.
Time $t_0 = m / \text{rate} = 39.47 / (5/6) \approx 47.36 \, s$.
The closest value among the options is $45 \, s$.

(B) Step $1$: Calculate the heat required to bring $100 \, g$ of ice from $-7^{\circ} C$ to $0^{\circ} C$.
$Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T = 100 \cdot 2.2 \cdot (0 - (-7)) = 100 \cdot 2.2 \cdot 7 = 1540 \, J$.
Step $2$: Calculate the heat released by $200 \, g$ of water cooling from $15^{\circ} C$ to $0^{\circ} C$.
$Q_2 = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T = 200 \cdot 4.2 \cdot (15 - 0) = 200 \cdot 4.2 \cdot 15 = 12600 \, J$.
Step $3$: Determine the heat available for melting the ice.
Heat available for melting = $Q_2 - Q_1 = 12600 - 1540 = 11060 \, J$.
Step $4$: Calculate the mass of ice that melts $(m_{\text{melt}})$.
$m_{\text{melt}} = \frac{\text{Heat available}}{L_f} = \frac{11060}{335} \approx 33.01 \, g$.
Step $5$: Calculate the remaining mass of ice.
$m_{\text{remaining}} = 100 - 33.01 = 66.99 \, g \approx 67 \, g$.

(A) Let the final temperature of the mixture be $T^{\circ} C$.
Heat lost by $2 \,kg$ of water at $90^{\circ} C$ to reach $T^{\circ} C$ is given by: $Q_{lost} = m_w s_w (90 - T) = 2 \times 4.18 \times (90 - T) = 8.36(90 - T) \,kJ$.
Heat gained by $1 \,kg$ of ice at $-20^{\circ} C$ to reach $0^{\circ} C$: $Q_1 = m_i s_i (0 - (-20)) = 1 \times 2.09 \times 20 = 41.8 \,kJ$.
Heat gained by $1 \,kg$ of ice at $0^{\circ} C$ to melt into water at $0^{\circ} C$: $Q_2 = m_i L_f = 1 \times 334.4 = 334.4 \,kJ$.
Heat gained by the resulting $1 \,kg$ of water at $0^{\circ} C$ to reach $T^{\circ} C$: $Q_3 = m_i s_w (T - 0) = 1 \times 4.18 \times T = 4.18T \,kJ$.
By the principle of calorimetry,$Q_{lost} = Q_1 + Q_2 + Q_3$.
$8.36(90 - T) = 41.8 + 334.4 + 4.18T$.
$752.4 - 8.36T = 376.2 + 4.18T$.
$752.4 - 376.2 = 4.18T + 8.36T$.
$376.2 = 12.54T$.
$T = \frac{376.2}{12.54} = 30^{\circ} C$.

(C) According to the principle of calorimetry,heat lost by the aluminium piece is equal to the heat gained by the water.
Let $m_a = 50 \,g = 0.05 \,kg$ be the mass of aluminium,$s_a = 900 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_a = (300^{\circ} C - 160^{\circ} C) = 140^{\circ} C$ be the change in temperature.
Let $m_w = 1 \,kg$ be the mass of water,$s_w = 4200 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_w = (T - 30^{\circ} C)$ be the change in temperature of water.
Heat lost by aluminium = $m_a s_a \Delta T_a = 0.05 \times 900 \times 140 = 6300 \,J$.
Heat gained by water = $m_w s_w \Delta T_w = 1 \times 4200 \times (T - 30) = 4200(T - 30)$.
Equating the two: $6300 = 4200(T - 30)$.
$T - 30 = \frac{6300}{4200} = 1.5$.
$T = 30 + 1.5 = 31.5^{\circ} C$.
Thus,the final temperature of the water is $31.5^{\circ} C$.

(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.
According to the principle of calorimetry,Heat lost by water = Heat gained by ice.
Heat lost by $100 \,g$ of water cooling from $80^{\circ} C$ to $0^{\circ} C$ is $Q_{lost} = m_w \cdot s_w \cdot \Delta T = 100 \times 1 \times (80 - 0) = 8000 \,cal$.
Heat required to melt $m$ grams of ice at $0^{\circ} C$ is $Q_{gained} = m \cdot L_f = m \times 80$.
Equating the two: $80m = 8000$.
Solving for $m$: $m = 100 \,g$.
Since the total mass of ice is $150 \,g$,the amount of ice that does not melt is $150 \,g - 100 \,g = 50 \,g$.

(B) When two blocks are brought into contact,the hotter block loses heat and the colder block gains heat.
Let $T_f$ be the final equilibrium temperature.
According to the principle of calorimetry,heat lost by the hotter block equals heat gained by the colder block.
$m \int_{T_f}^{80} s(T) dT = m \int_{20}^{T_f} s(T) dT$
Since the specific heat $s(T)$ is an increasing function of temperature,the average specific heat of the hotter block (in the range $T_f$ to $80^{\circ} C$) is greater than the average specific heat of the colder block (in the range $20^{\circ} C$ to $T_f$).
Let $s_{avg, hot}$ be the average specific heat of the hotter block and $s_{avg, cold}$ be the average specific heat of the colder block. Then $s_{avg, hot} > s_{avg, cold}$.
$s_{avg, hot} (80 - T_f) = s_{avg, cold} (T_f - 20)$
$\frac{80 - T_f}{T_f - 20} = \frac{s_{avg, cold}}{s_{avg, hot}} < 1$
$80 - T_f < T_f - 20$
$100 < 2T_f$
$T_f > 50^{\circ} C$
Therefore,the final temperature will be more than $50^{\circ} C$.

(C) Let the mass of steam required be $m \, g$.
Heat lost by steam at $100^{\circ} C$ to become water at $90^{\circ} C$:
$1$. Heat released during condensation: $Q_1 = m \times L_v = m \times 540 \, cal/g$ (using $L_v \approx 540 \, cal/g$ for steam).
$2$. Heat released by condensed water cooling from $100^{\circ} C$ to $90^{\circ} C$: $Q_2 = m \times c_w \times \Delta T = m \times 1 \times (100 - 90) = 10m \, cal$.
Total heat lost $= Q_1 + Q_2 = 540m + 10m = 550m \, cal$.
Heat gained by $100 \, g$ of water to raise its temperature from $24^{\circ} C$ to $90^{\circ} C$:
$Q_3 = m_{water} \times c_w \times \Delta T = 100 \times 1 \times (90 - 24) = 100 \times 66 = 6600 \, cal$.
By the principle of calorimetry,Heat Lost = Heat Gained:
$550m = 6600$
$m = \frac{6600}{550} = 12 \, g$.
Given the options,$13 \, g$ is the closest approximation.

(C) Given: Mass of ice $m_i = 10 \, g$, initial temperature $T_i = -20^{\circ} C$. Mass of water $m_w = 10 \, g$, initial temperature $T_w = 10^{\circ} C$. Specific heat of ice $c_i = 0.5 \, cal/g^{\circ} C$, specific heat of water $c_w = 1.0 \, cal/g^{\circ} C$. Latent heat of fusion $L_f = 80 \, cal/g$.
Heat released by water to reach $0^{\circ} C$: $Q_w = m_w c_w \Delta T = 10 \times 1 \times (10 - 0) = 100 \, cal$.
Heat required by ice to reach $0^{\circ} C$: $Q_{i1} = m_i c_i \Delta T = 10 \times 0.5 \times (0 - (-20)) = 100 \, cal$.
Since the heat released by water $(100 \, cal)$ is exactly equal to the heat required to bring the ice to $0^{\circ} C$ $(100 \, cal)$, no further heat is available to melt the ice.
Therefore, at equilibrium, the system remains at $0^{\circ} C$ with $10 \, g$ of ice and $10 \, g$ of water.

(B) The water equivalent $(W)$ of a body is defined as the mass of water that would have the same thermal capacity as the body.
Thermal capacity of the aluminium sample $= m \times c = 20 \,g \times 0.2 \,cal \,g^{-1} {}^{\circ}C^{-1} = 4 \,cal \,{}^{\circ}C^{-1}$.
Since the specific heat of water is $1 \,cal \,g^{-1} {}^{\circ}C^{-1}$,the water equivalent $W$ is given by:
$W = \frac{\text{Thermal Capacity}}{\text{Specific heat of water}} = \frac{4 \,cal \,{}^{\circ}C^{-1}}{1 \,cal \,g^{-1} {}^{\circ}C^{-1}} = 4 \,g$.
Therefore,the correct option is $B$.

(B) Heat released by water to reach $0^{\circ} C$: $\Delta Q_1 = m_w c_w \Delta T = 500 \times 1 \times (20 - 0) = 10,000 \,cal$.
Heat required by ice to reach $0^{\circ} C$: $\Delta Q_2 = m_i c_i \Delta T = 200 \times 0.5 \times (0 - (-20)) = 2000 \,cal$.
Remaining heat available to melt the ice: $\Delta Q_{rem} = \Delta Q_1 - \Delta Q_2 = 10,000 - 2,000 = 8,000 \,cal$.
Mass of ice melted: $m_{melted} = \frac{\Delta Q_{rem}}{L_f} = \frac{8,000}{80} = 100 \,g$.
Total mass of water in the vessel = Initial mass of water + Mass of melted ice = $500 + 100 = 600 \,g$.

(C) calorimeter is a device used to measure the amount of heat involved in a physical or chemical process.
It is typically made of a metal,most commonly copper,because metals have high thermal conductivity,which allows for rapid thermal equilibrium between the contents and the calorimeter.
Additionally,metals have a low specific heat capacity,which minimizes the heat absorbed by the calorimeter itself during the measurement process.
Therefore,the correct option is $C$.

(B) Let the mass of steam that condenses be $x \,g$.
The initial temperature of water is $288 \,K = 15^{\circ} C$.
The final temperature of water is $15^{\circ} C + 6.5^{\circ} C = 21.5^{\circ} C$.
Heat released by steam in condensing at $100^{\circ} C$ and cooling down to $21.5^{\circ} C$:
$Q_{\text{released}} = m_s L_v + m_s c_w \Delta T_s$
$Q_{\text{released}} = x \times 540 + x \times 1 \times (100 - 21.5) = 540x + 78.5x = 618.5x \,cal$.
Heat absorbed by water and calorimeter:
$Q_{\text{absorbed}} = (m_w c_w + W) \Delta T_w$
$Q_{\text{absorbed}} = (1100 \times 1 + 20) \times 6.5 = 1120 \times 6.5 = 7280 \,cal$.
By the principle of calorimetry,$Q_{\text{released}} = Q_{\text{absorbed}}$:
$618.5x = 7280$
$x = \frac{7280}{618.5} \approx 11.77 \,g$.
Rounding to the nearest provided option,$x = 11.7 \,g$.

(C) The heat energy required to raise the temperature of water is given by $Q = m \cdot s \cdot \Delta T$.
Here,$m = 2\,kg$,$s = 4200\,J\,kg^{-1}K^{-1}$,and $\Delta T = 60^{\circ}C - 10^{\circ}C = 50^{\circ}C$.
$Q = 2 \times 4200 \times 50 = 420,000\,J$.
The effective power delivered by the heater is $P_{eff} = \eta \times P = 0.70 \times 2000\,W = 1400\,W$.
The time required is $\Delta t = \frac{Q}{P_{eff}} = \frac{420,000}{1400} = 300\,s$.

(B) The total heat $Q = 420 \ J$ is used for two processes:
$1$. Raising the temperature of the entire mass $m$ (in grams) from $-5^{\circ}C$ to $0^{\circ}C$.
$2$. Melting $1 \ g$ of ice at $0^{\circ}C$.
Given:
Mass $m$ in grams $= m \times 10^{-3} \ kg$.
Specific heat of ice $c = 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$.
Latent heat of fusion $L = 3.36 \times 10^5 \ J \ kg^{-1}$.
Heat required to raise temperature: $Q_1 = m \times 10^{-3} \times 2100 \times (0 - (-5)) = m \times 2.1 \times 5 = 10.5m \ J$.
Heat required to melt $1 \ g$ of ice: $Q_2 = 1 \times 10^{-3} \times 3.36 \times 10^5 = 336 \ J$.
Total heat $Q = Q_1 + Q_2 = 420 \ J$.
$10.5m + 336 = 420$.
$10.5m = 420 - 336 = 84$.
$m = 84 / 10.5 = 8$.
Therefore,the value of $m$ is $8 \ g$.

(A) According to the principle of calorimetry,the heat lost by the hotter liquid is equal to the heat gained by the colder liquid.
Let $T_f$ be the final equilibrium temperature of the mixture.
Heat lost by the first liquid: $Q_{loss} = (2m) \cdot C \cdot (4T - T_f)$
Heat gained by the second liquid: $Q_{gain} = m \cdot (2C) \cdot (T_f - T)$
Equating the two: $(2m) \cdot C \cdot (4T - T_f) = m \cdot (2C) \cdot (T_f - T)$
Dividing both sides by $2mC$: $4T - T_f = T_f - T$
$4T + T = T_f + T_f$
$5T = 2T_f$
$T_f = \frac{5}{2} T$

(C) According to the principle of calorimetry,Heat lost by the copper = Heat gained by the water and calorimeter.
Let $S$ be the specific heat of copper.
Heat lost by copper = $m_{Cu} \cdot S \cdot \Delta T_{Cu} = 0.250 \ kg \cdot S \cdot (500^{\circ} C - 60^{\circ} C) = 0.250 \cdot S \cdot 440 = 110S$.
Heat gained by water and calorimeter = $(m_{water} + w) \cdot S_{water} \cdot \Delta T_{water}$,where $w$ is the water equivalent.
Heat gained = $(0.200 \ kg + 0.050 \ kg) \cdot 4200 \ J/kg^{\circ} C \cdot (60^{\circ} C - 20^{\circ} C) = 0.250 \cdot 4200 \cdot 40 = 42000 \ J$.
Equating the two: $110S = 42000$.
$S = 42000 / 110 \approx 381.8 \ J/kg^{\circ} C$.
Rounding to the nearest given option,$S \approx 380 \ J/kg^{\circ} C$.

(C) According to the principle of calorimetry,heat lost by the hot liquid equals the heat gained by the cold liquid.
$m_A s_A (T_A - T_f) = m_B s_B (T_f - T_B)$
Given ratios: $\frac{m_A}{m_B} = \frac{2}{3}$ and $\frac{s_A}{s_B} = \frac{3}{4}$.
Substituting the values: $\frac{2}{3} \times \frac{3}{4} \times (75 - T_f) = (T_f - 15)$
$\frac{1}{2} \times (75 - T_f) = T_f - 15$
$75 - T_f = 2(T_f - 15)$
$75 - T_f = 2T_f - 30$
$3T_f = 105$
$T_f = 35^{\circ} C$.

(C) Heat released by $4 \text{ g}$ of steam to condense into water at $100^{\circ} C$ is $Q_1 = m_s L = 4 \times 540 = 2160 \text{ cal}$.
Heat required by $20 \text{ g}$ of water to reach $100^{\circ} C$ is $Q_2 = m_w c \Delta T = 20 \times 1 \times (100 - 46) = 20 \times 54 = 1080 \text{ cal}$.
Since $Q_1 > Q_2$, the steam will not condense completely.
The amount of steam condensed is $m' = \frac{Q_2}{L} = \frac{1080}{540} = 2 \text{ g}$.
Final mass of water in the container = Initial mass of water + Mass of condensed steam = $20 \text{ g} + 2 \text{ g} = 22 \text{ g}$.

(C) According to the principle of calorimetry,the heat lost by the hotter liquid is equal to the heat gained by the colder liquid.
Let $m$ be the mass of each liquid and $X$ be the specific heat of the second liquid.
Heat lost by the first liquid = $m \times 0.8 \times (60 - 53)$
Heat gained by the second liquid = $m \times X \times (53 - 45)$
Equating the two:
$m \times 0.8 \times 7 = m \times X \times 8$
$5.6 = 8X$
$X = \frac{5.6}{8} = 0.7 \ cal / g^{\circ} C$

(A) According to the principle of calorimetry,heat lost by the hot body equals heat gained by the cold body and the calorimeter.
Let $w$ be the water equivalent of the calorimeter.
Heat lost by $0.4 \ kg$ of water at $30^{\circ} C$ cooling to $27^{\circ} C$ is:
$Q_{lost} = m_1 c \Delta T_1 = 0.4 \times 1 \times (30 - 27) = 0.4 \times 3 = 1.2 \ kcal$ (considering specific heat of water $c = 1 \ kcal/kg^{\circ}C$)
Heat gained by $0.15 \ kg$ of water at $25^{\circ} C$ heating to $27^{\circ} C$ is:
$Q_{gained, water} = m_2 c \Delta T_2 = 0.15 \times 1 \times (27 - 25) = 0.15 \times 2 = 0.3 \ kcal$
Heat gained by the calorimeter is:
$Q_{gained, cal} = w \times c \times \Delta T_2 = w \times 1 \times (27 - 25) = 2w$
Equating heat lost and heat gained:
$1.2 = 0.3 + 2w$
$0.9 = 2w$
$w = 0.45 \ kg$

(C) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.
Heat lost by metal = Heat gained by water.
$m_{\text{metal}} \times s_{\text{metal}} \times \Delta T_{\text{metal}} = m_{\text{water}} \times s_{\text{water}} \times \Delta T_{\text{water}}$
Given: $m_{\text{metal}} = 100 \,g = 0.1 \,kg$,$T_{\text{initial, metal}} = 80^{\circ} C$,$T_{\text{final}} = 15.69^{\circ} C$,$m_{\text{water}} = 1 \,kg = 1000 \,g$,$T_{\text{initial, water}} = 15^{\circ} C$,$s_{\text{water}} = 4.18 \,J/g^{\circ} C$.
Substituting the values:
$100 \times s_{\text{metal}} \times (80 - 15.69) = 1000 \times 4.18 \times (15.69 - 15)$
$100 \times s_{\text{metal}} \times 64.31 = 1000 \times 4.18 \times 0.69$
$6431 \times s_{\text{metal}} = 2884.2$
$s_{\text{metal}} = \frac{2884.2}{6431} \approx 0.4485 \,J/g^{\circ} C \approx 0.45 \,J/g^{\circ} C$.

(B) Let the final temperature of the mixture be $T$.
Heat gained by ice = Heat lost by water.
Heat gained by ice = (Heat to melt ice) + (Heat to raise temperature of melted ice from $0^{\circ}C$ to $T$).
$Q_{gain} = m_i L_f + m_i S_w (T - 0)$
$Q_{gain} = (0.1 \text{ kg} \times 3.36 \times 10^5 \text{ J/kg}) + (0.1 \text{ kg} \times 4.2 \times 10^3 \text{ J/kg K} \times T)$
$Q_{gain} = 33600 + 420 T$
Heat lost by water = $m_w S_w (100 - T)$
$Q_{lost} = 0.1 \text{ kg} \times 4.2 \times 10^3 \text{ J/kg K} \times (100 - T)$
$Q_{lost} = 420 (100 - T) = 42000 - 420 T$
Equating heat gained and lost:
$33600 + 420 T = 42000 - 420 T$
$840 T = 8400$
$T = 10^{\circ}C$.

(A) Let the final temperature of the mixture be $t$.
Heat lost by water at $80^{\circ} C = m_1 s \Delta t_1 = (V_1 \rho) s (80^{\circ} - t)$.
Heat gained by water at $60^{\circ} C = m_2 s \Delta t_2 = (V_2 \rho) s (t - 60^{\circ})$.
According to the principle of calorimetry,Heat lost = Heat gained.
Since the density $\rho$ and specific heat $s$ are the same for both:
$V_1 (80^{\circ} - t) = V_2 (t - 60^{\circ})$.
Substituting the values:
$0.1 (80 - t) = 0.3 (t - 60)$.
$8 - 0.1t = 0.3t - 18$.
$26 = 0.4t$.
$t = \frac{26}{0.4} = 65^{\circ} C$.

(B) Power of the machine $P = 10 \,kW = 10,000 \,W$.
Time $t = 3 \,minutes = 3 \times 60 = 180 \,s$.
Total energy produced $E = P \times t = 10,000 \times 180 = 1,800,000 \,J$.
Heat absorbed by the block $Q = 50 \% \text{ of } E = 0.5 \times 1,800,000 = 900,000 \,J$.
Using the formula $Q = mc\Delta T$, where $m = 25 \,kg$ and $c = 900 \,J \,kg^{-1} \,K^{-1}$:
$900,000 = 25 \times 900 \times \Delta T$.
$900,000 = 22,500 \times \Delta T$.
$\Delta T = \frac{900,000}{22,500} = 40^{\circ} C$.
Therefore, the rise in temperature is $40^{\circ} C$.

(A) Let $m$ be the mass of steam condensed into water.
Heat lost by steam = Heat gained by water.
Heat lost by steam = (Heat released during condensation) + (Heat released by condensed water cooling from $100^{\circ} C$ to $70^{\circ} C$)
$Q_{lost} = m \times L + m \times c \times \Delta T$
$Q_{lost} = m \times 540 + m \times 1 \times (100 - 70) = 540m + 30m = 570m$
Heat gained by water = $m_{water} \times c \times \Delta T$
$Q_{gained} = 114 \times 1 \times (70 - 30) = 114 \times 40 = 4560 \ cal$
Equating heat lost and gained: $570m = 4560$
$m = 4560 / 570 = 8 \ g$
The total mass of water in the mixture = (Initial mass of water) + (Mass of condensed steam)
Total mass = $114 \ g + 8 \ g = 122 \ g$.

(D) According to the principle of calorimetry,the heat lost by the water is equal to the heat gained by the ice to melt.
Heat lost by water $(Q_{lost})$ = $m \cdot s \cdot \Delta T = m \cdot s \cdot \theta$.
Heat gained by ice to melt $(Q_{gained})$ = $m_{melted} \cdot L$.
Equating the two: $m \cdot s \cdot \theta = m_{melted} \cdot L$.
Therefore,the mass of ice melted $(m_{melted})$ = $\frac{ms \theta}{L}$.

(B) Let $m_s$ be the mass of the steam added.
According to the principle of calorimetry,Heat lost by steam = Heat gained by water.
Heat lost by steam consists of two parts:
$1$. Heat released during condensation of steam at $100^{\circ} C$ to water at $100^{\circ} C$: $Q_1 = m_s \times L = m_s \times 540$.
$2$. Heat released by the condensed water cooling from $100^{\circ} C$ to $90^{\circ} C$: $Q_2 = m_s \times c \times \Delta T = m_s \times 1 \times (100 - 90) = 10 m_s$.
Total heat lost = $540 m_s + 10 m_s = 550 m_s$.
Heat gained by $100 \ g$ of water to raise its temperature from $24^{\circ} C$ to $90^{\circ} C$:
$Q_{gain} = m_w \times c \times \Delta T = 100 \times 1 \times (90 - 24) = 100 \times 66 = 6600 \ cal$.
Equating heat lost and gained: $550 m_s = 6600$.
$m_s = 6600 / 550 = 12 \ g$.

(C) According to the principle of calorimetry,the heat gained by the metal ball is equal to the heat lost by the water.
Let $m_1 = 100 \ g$ be the mass of the metal,$s_1$ be its specific heat,and $T_1 = 20^{\circ} C$ be its initial temperature.
Let $m_2 = 200 \ g$ (since density of water is $1 \ g/ml$) be the mass of water,$s_2$ be its specific heat,and $T_2 = 80^{\circ} C$ be its initial temperature.
The final equilibrium temperature is $T = 70^{\circ} C$.
Heat gained by metal = $m_1 s_1 (T - T_1) = 100 \times s_1 \times (70 - 20) = 5000 s_1$.
Heat lost by water = $m_2 s_2 (T_2 - T) = 200 \times s_2 \times (80 - 70) = 2000 s_2$.
Equating the two: $5000 s_1 = 2000 s_2$.
Therefore,the ratio $\frac{s_1}{s_2} = \frac{2000}{5000} = \frac{2}{5}$.

(B) According to the principle of calorimetry,Heat lost by water = Heat gained by ice.
Let $m$ be the mass of water in grams.
Heat lost by water $= m \times c_w \times (T_i - T_f) = m \times 1 \times (30 - 6) = 24m \text{ cal}$.
Heat gained by ice consists of three parts:
$1$. Heating ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_i \times c_i \times \Delta T = 5 \times 0.5 \times (0 - (-20)) = 5 \times 0.5 \times 20 = 50 \text{ cal}$.
$2$. Melting ice at $0^{\circ} C$: $Q_2 = m_i \times L_f = 5 \times 80 = 400 \text{ cal}$.
$3$. Heating the melted water from $0^{\circ} C$ to $6^{\circ} C$: $Q_3 = m_i \times c_w \times \Delta T = 5 \times 1 \times (6 - 0) = 30 \text{ cal}$.
Total heat gained $= 50 + 400 + 30 = 480 \text{ cal}$.
Equating heat lost and gained: $24m = 480$.
$m = 480 / 24 = 20 \text{ g}$.

(B) Step $1$: Calculate heat required to bring ice from $-10^{\circ} C$ to $0^{\circ} C$ (ice): $Q_1 = m_i c_i \Delta T = 50 \times 0.5 \times 10 = 250 \,cal$.
Step $2$: Calculate heat required to melt $50 \,g$ of ice at $0^{\circ} C$: $Q_2 = m_i L_f = 50 \times 80 = 4000 \,cal$.
Total heat required by ice to become water at $0^{\circ} C$ is $Q_{total} = 250 + 4000 = 4250 \,cal$.
Step $3$: Calculate heat released by $200 \,g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{release} = m_w c_w \Delta T = 200 \times 1 \times 30 = 6000 \,cal$.
Since $Q_{release} > Q_{total}$,the ice melts completely and the final temperature $T_f$ will be above $0^{\circ} C$.
Step $4$: Apply the principle of calorimetry: Heat lost by water = Heat gained by ice.
$200 \times 1 \times (30 - T_f) = 4250 + 50 \times 1 \times (T_f - 0)$.
$6000 - 200 T_f = 4250 + 50 T_f$.
$1750 = 250 T_f$.
$T_f = 7^{\circ} C$.

(D) Given: Mass of ice $m_{\text{ice}} = 5 \ g$,initial temperature $T_{\text{ice}} = -30^{\circ} C$. Mass of water $m_w = 20 \ g$,initial temperature $T_w = 35^{\circ} C$.
Specific heat of ice $s_{\text{ice}} = 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $s_w = 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion $L_f = 80 \ cal \ g^{-1}$.
Heat required to raise ice from $-30^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{\text{ice}} s_{\text{ice}} \Delta T = 5 \times 0.5 \times 30 = 75 \ cal$.
Heat required to melt ice at $0^{\circ} C$: $Q_2 = m_{\text{ice}} L_f = 5 \times 80 = 400 \ cal$.
Total heat required to convert ice at $-30^{\circ} C$ to water at $0^{\circ} C$ is $Q_{\text{total}} = 75 + 400 = 475 \ cal$.
Heat available from water cooling from $35^{\circ} C$ to $0^{\circ} C$: $Q_{\text{avail}} = m_w s_w \Delta T = 20 \times 1 \times 35 = 700 \ cal$.
Since $Q_{\text{avail}} > Q_{\text{total}}$,the final temperature $T$ will be above $0^{\circ} C$.
Heat lost by water = Heat gained by ice:
$m_w s_w (35 - T) = Q_{\text{total}} + m_{\text{ice}} s_w (T - 0)$
$20 \times 1 \times (35 - T) = 475 + 5 \times 1 \times T$
$700 - 20T = 475 + 5T$
$225 = 25T$
$T = 9^{\circ} C$.

(C) Given: Mass of metal block,$m = 3.3 \ kg = 3300 \ g$.
Temperature change,$\Delta T = 400^{\circ} C$.
Specific heat of metal,$s_m = 0.4 \ J \ g^{-1} \ K^{-1}$.
Latent heat of fusion of ice,$L_f = 330 \ J \ g^{-1}$.
The heat released by the metal block is $Q = m \cdot s_m \cdot \Delta T$.
$Q = 3300 \ g \times 0.4 \ J \ g^{-1} \ K^{-1} \times 400 \ K = 528,000 \ J$.
Let $m'$ be the mass of ice melted. The heat required to melt the ice is $Q = m' \cdot L_f$.
$m' = \frac{Q}{L_f} = \frac{528,000 \ J}{330 \ J \ g^{-1}} = 1600 \ g$.
Therefore,$m' = 1.6 \ kg$.

(B) Let the water equivalent of the calorimeter be $W$ (in $kg$).
According to the principle of calorimetry, the heat lost by the hot water is equal to the heat gained by the cold water and the calorimeter.
Heat lost by hot water = $m_2 c (T_2 - T_f)$
Heat gained by cold water = $m_1 c (T_f - T_1)$
Heat gained by calorimeter = $W c (T_f - T_1)$
Here, $m_1 = 0.5 \,kg$, $T_1 = 30^{\circ} C$, $m_2 = 0.3 \,kg$, $T_2 = 60^{\circ} C$, and $T_f = 40^{\circ} C$.
Since the specific heat capacity $c$ of water is common to all terms, it cancels out:
$m_2 (T_2 - T_f) = (m_1 + W) (T_f - T_1)$
$0.3 (60 - 40) = (0.5 + W) (40 - 30)$
$0.3 \times 20 = (0.5 + W) \times 10$
$6 = 5 + 10W$
$1 = 10W$
$W = 0.1 \,kg$.

(B) Let the final equilibrium temperature of the mixture be $T^{\circ} C$.
According to the principle of calorimetry,heat gained by the cold water equals the heat lost by the hot water.
Heat gained by $50 \ g$ of water at $10^{\circ} C$: $H_{\text{gain}} = m c \Delta T = 50 \cdot c \cdot (T - 10)$.
Heat lost by $50 \ g$ of water at $100^{\circ} C$: $H_{\text{loss}} = m c \Delta T = 50 \cdot c \cdot (100 - T)$.
Equating the two: $50 \cdot c \cdot (T - 10) = 50 \cdot c \cdot (100 - T)$.
Dividing both sides by $50 \cdot c$: $T - 10 = 100 - T$.
$2T = 110$.
$T = 55^{\circ} C$.

(C) Mass of steam,$m_s = 1 \text{ kg} = 1000 \text{ g} = 10^3 \text{ g}$.
Initial temperature of steam,$T_i = 150^{\circ} \text{C}$.
Final temperature of steam (as water),$T_f = 90^{\circ} \text{C}$.
Latent heat of steam,$L_v = 540 \text{ cal g}^{-1}$.
Specific heat of steam,$c_s = 1 \text{ cal g}^{-1} {}^{\circ} \text{C}^{-1}$.
Specific heat of water,$c_w = 1 \text{ cal g}^{-1} {}^{\circ} \text{C}^{-1}$.
Heat lost by steam $(Q_{lost})$ consists of cooling steam from $150^{\circ} \text{C}$ to $100^{\circ} \text{C}$,condensation at $100^{\circ} \text{C}$,and cooling water from $100^{\circ} \text{C}$ to $90^{\circ} \text{C}$.
$Q_{lost} = m_s c_s (150-100) + m_s L_v + m_s c_w (100-90)$
$Q_{lost} = 1000 \times 1 \times 50 + 1000 \times 540 + 1000 \times 1 \times 10$
$Q_{lost} = 50,000 + 540,000 + 10,000 = 600,000 \text{ cal} = 60 \times 10^4 \text{ cal}$.
Heat gained by $20 \text{ L}$ of water $(Q_{gained})$: $m_w = 20 \text{ kg} = 20,000 \text{ g}$.
$Q_{gained} = m_w c_w \Delta T = 20,000 \times 1 \times \Delta T = 20,000 \Delta T$.
By the principle of calorimetry,$Q_{lost} = Q_{gained}$.
$600,000 = 20,000 \Delta T$.
$\Delta T = \frac{600,000}{20,000} = 30^{\circ} \text{C}$.