Principle of Calorimetry and Water Equivalent Questions in English

(C) The power $P$ supplied by the geyser is used to heat the water flowing through it.
The formula for heat transfer is $P = \frac{dm}{dt} c \Delta T$, where $\frac{dm}{dt}$ is the mass flow rate, $c$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature.
Given: Power $P = 2100 \, W$, mass flow rate $\frac{dm}{dt} = 20 \, g/s = 0.02 \, kg/s$, specific heat capacity of water $c = 4200 \, J/(kg \cdot ^{\circ}C)$, and inlet temperature $T_{in} = 10^{\circ}C$.
Substituting the values: $2100 = 0.02 \times 4200 \times (T_{out} - 10)$.
$2100 = 84 \times (T_{out} - 10)$.
$T_{out} - 10 = \frac{2100}{84} = 25$.
$T_{out} = 25 + 10 = 35^{\circ}C$.

(A) Let $W$ be the water equivalent of the flask in grams and $L$ be the latent heat of fusion of ice in $cal/g$.
Step $1$: Heat lost by water and flask = Heat gained by ice.
$(200 + W)(70 - 40) = 50L + 50(40 - 0)$
$30(200 + W) = 50L + 2000$
$6000 + 30W = 50L + 2000 \implies 50L - 30W = 4000 \implies 5L - 3W = 400$ ---$(1)$
Step $2$: For the second addition,total mass of water is $(200 + 50 + 80) = 330\,g$.
Heat lost by water and flask = Heat gained by ice.
$(200 + 50 + W)(40 - 10) = 80L + 80(10 - 0)$
$30(250 + W) = 80L + 800$
$7500 + 30W = 80L + 800 \implies 80L - 30W = 6700 \implies 8L - 3W = 670$ ---$(2)$
Step $3$: Subtract $(1)$ from $(2)$:
$(8L - 3W) - (5L - 3W) = 670 - 400$
$3L = 270 \implies L = 90\,cal/g$.
Step $4$: Convert to $J/kg$:
$L = 90 \times 4.184 \times 10^3\,J/kg \approx 3.76 \times 10^5\,J/kg \approx 3.8 \times 10^5\,J/kg$.

(B) Let $M$ be the total mass of the supercooled water.
Let $m$ be the mass of water that freezes into ice.
The latent heat released by the freezing of mass $m$ is $Q_{released} = m \times L_f$,where $L_f = 80 \ cal/g$.
This heat is absorbed by the remaining water to raise its temperature from $-15^{\circ} C$ to $0^{\circ} C$.
The heat required is $Q_{absorbed} = M \times c_w \times \Delta T$,where $c_w = 1 \ cal/g^{\circ} C$ and $\Delta T = 15^{\circ} C$.
Since the flask is thermally insulated,$Q_{released} = Q_{absorbed}$.
$80m = M \times 1 \times 15$.
$\frac{m}{M} = \frac{15}{80} = \frac{3}{16}$.
Wait,re-evaluating the specific heat of ice: The heat released by freezing $m$ grams of water is $m \times 80$. This heat warms the entire mass $M$ from $-15^{\circ} C$ to $0^{\circ} C$. Assuming specific heat of water is $1 \ cal/g^{\circ} C$:
$80m = M \times 1 \times 15 \implies m/M = 15/80 = 3/16$.
However,if we consider the specific heat of ice as $0.5 \ cal/g^{\circ} C$ for the initial state,the equation is $80m = M \times 0.5 \times 15$.
$80m = 7.5M \implies m/M = 7.5/80 = 15/160 = 3/32$.
Given the provided options,the standard calculation used in such problems is $80m = M \times 1 \times 15$ or similar. Re-checking the provided solution logic: $80m = m \times 0.5 \times 15 + (M-m) \times 1 \times 15$. This assumes the ice formed stays at $0^{\circ} C$ and the rest of the water warms up.
$80m = 7.5m + 15M - 15m \implies 87.5m = 15M \implies m/M = 15/87.5 = 6/35$.

(A) Let $m$ be the mass of steam that condenses in $kg$.
Heat lost by steam = Heat gained by water and calorimeter.
Heat lost by steam = $m L_v + m c_w (T_{steam} - T_{final})$,where $L_v = 2260 \, kJ/kg$ and $c_w = 4.18 \, kJ/kg \cdot ^oC$.
Heat lost = $m(2260) + m(4.18)(100 - 80) = m(2260 + 83.6) = 2343.6 m$.
Heat gained by water and calorimeter = $(m_{water} + m_{eq}) c_w (T_{final} - T_{initial})$.
Heat gained = $(1.1 + 0.02) \times 4.18 \times (80 - 15) = 1.12 \times 4.18 \times 65 = 304.304 \, kJ$.
Equating heat lost and gained: $2343.6 m = 304.304$.
$m = 304.304 / 2343.6 \approx 0.1298 \, kg \approx 0.130 \, kg$.

(B) According to the principle of calorimetry,heat lost by the copper ball is equal to the heat gained by the copper calorimeter and the water.
Let $m_b = 100 \ gm$ be the mass of the copper ball,$m_c = 100 \ gm$ be the mass of the calorimeter,and $m_w = 170 \ gm$ be the mass of water.
Specific heat of copper $s_{Cu} = 0.1 \ cal/gm ^\circ C$ and specific heat of water $s_w = 1 \ cal/gm ^\circ C$.
Initial temperature of the ball is $T$,and the final equilibrium temperature $T_f = 75 ^\circ C$. Initial temperature of the calorimeter and water $T_0 = 30 ^\circ C$.
Heat lost by the ball $= m_b s_{Cu} (T - T_f) = 100 \times 0.1 \times (T - 75) = 10(T - 75)$.
Heat gained by the calorimeter $= m_c s_{Cu} (T_f - T_0) = 100 \times 0.1 \times (75 - 30) = 10 \times 45 = 450 \ cal$.
Heat gained by the water $= m_w s_w (T_f - T_0) = 170 \times 1 \times (75 - 30) = 170 \times 45 = 7650 \ cal$.
Equating the heat lost and gained:
$10(T - 75) = 450 + 7650$
$10T - 750 = 8100$
$10T = 8850$
$T = 885 ^\circ C$.

(B) The rate of heat removal is given by $Q/t = 6.7 \times 10^9 \ J/min$.
To convert this to $J/s$,we divide by $60$: $\frac{dQ}{dt} = \frac{6.7 \times 10^9}{60} \ J/s$.
The formula for heat transfer is $\frac{dQ}{dt} = \frac{dm}{dt} \times S \times \Delta T$,where $\frac{dm}{dt}$ is the rate of water flow.
Given $S = 4200 \ J/kg \ ^oC$ and $\Delta T = 14.0 \ ^oC - 6.0 \ ^oC = 8.0 \ ^oC$.
Substituting the values: $\frac{6.7 \times 10^9}{60} = \frac{dm}{dt} \times 4200 \times 8$.
Rearranging for the rate of water flow: $\frac{dm}{dt} = \frac{6.7 \times 10^9}{4200 \times 8 \times 60} \ kg \ s^{-1}$.

(B) Let $m$ be the mass of both alcohol and water. Let $S_A$ and $S_W$ be the specific heats of alcohol and water,respectively. We are given $S_A \approx 0.5 S_W$,which implies $S_W > S_A$.
When mixed,the heat lost by the hotter substance equals the heat gained by the colder substance: $m S_A (T_f - T_A) = m S_W (T_W - T_f)$ (assuming $T_W > T_A$).
Simplifying,$S_A T_f - S_A T_A = S_W T_W - S_W T_f$.
$T_f (S_A + S_W) = S_W T_W + S_A T_A$.
$T_f = \frac{S_W T_W + S_A T_A}{S_W + S_A}$.
Since $S_W > S_A$,the final temperature $T_f$ will be weighted more towards the temperature of the substance with the higher specific heat,which is water $(T_W)$.

(B) According to the principle of calorimetry,the total heat lost by the hotter bodies must equal the total heat gained by the colder bodies,assuming no heat loss to the surroundings.
Since all blocks are made of copper,they have the same specific heat capacity $s$.
The heat exchange equation is given by:
$M_1 s(T_1 - T) + M_2 s(T_2 - T) + M_3 s(T_3 - T) = 0$
Dividing by $s$ (as $s \neq 0$):
$M_1(T_1 - T) + M_2(T_2 - T) + M_3(T_3 - T) = 0$
$M_1 T_1 - M_1 T + M_2 T_2 - M_2 T + M_3 T_3 - M_3 T = 0$
$M_1 T_1 + M_2 T_2 + M_3 T_3 = T(M_1 + M_2 + M_3)$
$T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3}$

(A) At equilibrium,the temperatures of both objects are the same,let this be $T$.
Since the system is isolated,the total heat exchange is zero.
Heat gained by $A$ = Heat lost by $B$
$m_A c_A (T - T_A) = m_B c_B (T_B - T)$
Given $m_A = m_B = 1.0 \ kg$,$T_A = 0^{\circ}C$,$T_B = 100^{\circ}C$,and $c_A < c_B$.
$c_A (T - 0) = c_B (100 - T)$
$c_A T = 100 c_B - c_B T$
$T(c_A + c_B) = 100 c_B$
$T = \frac{100 c_B}{c_A + c_B}$
Since $c_A < c_B$,we have $c_A + c_B < 2c_B$.
Therefore,$\frac{100 c_B}{c_A + c_B} > \frac{100 c_B}{2 c_B} = 50^{\circ}C$.
Thus,the final equilibrium temperature $T_A = T_B > 50^{\circ}C$.

(A) Let $m$ be the mass of steam in grams. The density of water is $1 \text{ g/cm}^3$,so $6 \text{ litres} = 6000 \text{ g}$.
Heat gained by water to reach $40^{\circ}C$: $Q_1 = m_w c_w \Delta T_w = 6000 \times 1 \times (40 - 20) = 120,000 \text{ cal}$.
Heat lost by steam at $100^{\circ}C$ to become water at $40^{\circ}C$: $Q_2 = m L_v + m c_w \Delta T_s = m \times 540 + m \times 1 \times (100 - 40) = 540m + 60m = 600m$.
Equating heat gained and lost: $120,000 = 600m$.
Solving for $m$: $m = \frac{120,000}{600} = 200 \text{ g}$.

(B) The final equilibrium temperature $T$ of a mixture of two substances with masses $m_1, m_2$,specific heats $S_1, S_2$,and initial temperatures $T_1, T_2$ is given by:
$T = \frac{m_1 S_1 T_1 + m_2 S_2 T_2}{m_1 S_1 + m_2 S_2}$
Given that the masses are identical $(m_A = m_W = m)$:
$T = \frac{m S_A T_A + m S_W T_W}{m S_A + m S_W} = \frac{S_A T_A + S_W T_W}{S_A + S_W}$
Given that the specific heat of alcohol is half that of water $(S_A = \frac{1}{2} S_W)$:
$T = \frac{(\frac{1}{2} S_W) T_A + S_W T_W}{\frac{1}{2} S_W + S_W} = \frac{\frac{1}{2} T_A + T_W}{\frac{3}{2}} = \frac{T_A + 2 T_W}{3}$
$T = \frac{1}{3} T_A + \frac{2}{3} T_W$
Since the weight of $T_W$ is twice that of $T_A$ in the final expression,the final temperature $T$ is closer to $T_W$ than to $T_A$.

(C) Heat released when $x \text{ grams}$ of steam at $100^{\circ}C$ condenses into water at $100^{\circ}C$ is given by $H_1 = m L_v = x \times 540 \text{ cal}$.
Heat absorbed by $y \text{ grams}$ of ice at $0^{\circ}C$ to turn into water at $100^{\circ}C$ consists of two parts: melting the ice and heating the resulting water.
$H_2 = m L_f + m c \Delta T = y \times 80 + y \times 1 \times (100 - 0) = 80y + 100y = 180y \text{ cal}$.
According to the principle of calorimetry,heat lost equals heat gained:
$H_1 = H_2$
$540x = 180y$
Rearranging to find the ratio $y/x$:
$\frac{y}{x} = \frac{540}{180} = 3$.

(A) Heat lost by water at $80 \ ^\circ C$ to reach $0 \ ^\circ C$ is given by $Q_{lost} = m_w c_w \Delta T = 540 \times 1 \times (80 - 0) = 43200 \ cal$.
Heat required to melt $540 \ g$ of ice at $0 \ ^\circ C$ is $Q_{req} = m_i L_f = 540 \times 80 = 43200 \ cal$.
Since the heat lost by water is exactly equal to the heat required to melt the ice,the entire ice will melt at $0 \ ^\circ C$ and the final temperature of the mixture will be $0 \ ^\circ C$.

(C) Let $m_m = 5 \ kg = 5000 \ g$ be the mass of the metal,$T_m = -50 \ ^\circ C$ be its initial temperature,and $c_m$ be its specific heat.
Let $m_w = 20 \ kg = 20000 \ g$ be the mass of water,$T_w = 52 \ ^\circ C$ be its initial temperature,and $c_w = 1 \ Cal/g \ ^\circ C$ be the specific heat of water.
The final equilibrium temperature $T_f = 52 \ ^\circ C - 2 \ ^\circ C = 50 \ ^\circ C$.
According to the principle of calorimetry,Heat lost by water = Heat gained by metal.
$m_w c_w (T_w - T_f) = m_m c_m (T_f - T_m)$
$20000 \times 1 \times (52 - 50) = 5000 \times c_m \times (50 - (-50))$
$20000 \times 2 = 5000 \times c_m \times 100$
$40000 = 500000 \times c_m$
$c_m = \frac{40000}{500000} = \frac{4}{50} = 0.08 \ Cal/g \ ^\circ C$.

(C) According to the principle of calorimetry,the heat lost by the copper block is equal to the heat gained by the ice to melt.
Heat lost by copper $(Q_{lost})$ = $m_{Cu} \cdot c_{Cu} \cdot \Delta T$
$Q_{lost} = 2\, kg \times 400\, J/kg/ ^\circ C \times (500^\circ C - 0^\circ C) = 400,000\, J$
Heat gained by ice $(Q_{gained})$ = $m_{ice} \cdot L_f$
$Q_{gained} = m \times 3.5 \times 10^5\, J/kg$
Equating the two: $m \times 3.5 \times 10^5 = 400,000$
$m = \frac{400,000}{350,000} = \frac{40}{35} = \frac{8}{7}\, kg$
Thus,the amount of ice that melts is $8/7\, kg$.

(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 phase change) + (Heat released by condensed water cooling from $100\,^{\circ}C$ to $90\,^{\circ}C$).
Heat lost = $m \times L_v + m \times c_w \times \Delta T_1 = m \times 540 + m \times 1 \times (100 - 90) = 540m + 10m = 550m$.
Heat gained by water = $m_{water} \times c_w \times \Delta T_2 = 22 \times 1 \times (90 - 20) = 22 \times 70 = 1540\,cal$.
Equating heat lost and heat gained: $550m = 1540$.
$m = \frac{1540}{550} = 2.8\,g$.
The total mass of water present = (Initial mass of water) + (Mass of condensed steam) = $22\,g + 2.8\,g = 24.8\,g$.

(A) According to the principle of calorimetry,heat lost by the metal equals heat gained by the water.
Let $m_1 = 2\,kg$ be the mass of the metal and $s_1$ be its specific heat capacity.
Let $m_2 = 1\,kg$ be the mass of the water and $s_2$ be its specific heat capacity.
Given $s_1 = \frac{1}{2} s_2$.
Let $\theta$ be the final equilibrium temperature.
Heat lost by metal $= m_1 s_1 (100 - \theta) = 2 \times (\frac{s_2}{2}) \times (100 - \theta) = s_2 (100 - \theta)$.
Heat gained by water $= m_2 s_2 (\theta - 0) = 1 \times s_2 \times \theta = s_2 \theta$.
Equating the two: $s_2 (100 - \theta) = s_2 \theta$.
$100 - \theta = \theta$.
$2\theta = 100$.
$\theta = 50\,^{\circ}C$.

(D) To melt the ice completely,the heat lost by the water must be equal to the heat gained by the ice to undergo a phase change from solid to liquid at $0 \, ^oC$.
Heat gained by $10 \, g$ of ice to melt: $Q_{gain} = m_{ice} \cdot L_f = 10 \, g \times 80 \, cal/g = 800 \, cal$.
Heat lost by $m \, g$ of water cooling from $50 \, ^oC$ to $0 \, ^oC$: $Q_{lost} = m \cdot S_W \cdot \Delta T = m \times 1 \, cal/g \cdot ^oC \times (50 \, ^oC - 0 \, ^oC) = 50m \, cal$.
Equating the heat gained and lost: $800 = 50m$.
Solving for $m$: $m = \frac{800}{50} = 16 \, g$.

(D) Let the final equilibrium temperature be $\theta$.
According to the principle of calorimetry,Heat gained by the cold system = Heat lost by the hot water.
Heat gained by the vessel and the initial water: $(m_{vessel} \cdot c + m_{water1} \cdot c) \cdot (\theta - T_{initial}) = (10 + 110) \cdot 1 \cdot (\theta - 10) = 120(\theta - 10)$.
Heat lost by the added hot water: $m_{water2} \cdot c \cdot (T_{hot} - \theta) = 220 \cdot 1 \cdot (70 - \theta) = 220(70 - \theta)$.
Equating the two: $120(\theta - 10) = 220(70 - \theta)$.
Dividing by $20$: $6(\theta - 10) = 11(70 - \theta)$.
$6\theta - 60 = 770 - 11\theta$.
$17\theta = 830$.
$\theta = \frac{830}{17} \approx 48.82^{\circ}C$.
Rounding to the nearest integer,the final temperature is approximately $49^{\circ}C$.

(A) Let the mass of water taken from tank $A$ be $x$ kg.
Then, the mass of water taken from tank $B$ is $(40-x)$ kg.
According to the principle of calorimetry, heat gained by cold water equals heat lost by hot water.
$m_A s (T_f - T_A) = m_B s (T_B - T_f)$
Where $m_A = x$, $m_B = (40-x)$, $T_A = 30,^{\circ}C$, $T_B = 80,^{\circ}C$, and $T_f = 50,^{\circ}C$.
Since the specific heat $s$ is the same for both, it cancels out:
$x(50 - 30) = (40 - x)(80 - 50)$
$20x = 30(40 - x)$
$20x = 1200 - 30x$
$50x = 1200$
$x = 24$ kg.
Thus, mass from tank $A = 24$ kg and mass from tank $B = 40 - 24 = 16$ kg.

(A) According to the principle of calorimetry,heat lost by the hot body equals heat gained by the cold body.
Let the final temperature be $\theta$.
Heat gained by ice to melt at $0\,^{\circ}C$ is $Q_1 = m_i L_f = 5 \times 80 = 400\,cal$.
Heat gained by the melted ice to reach temperature $\theta$ is $Q_2 = m_i c_w (\theta - 0) = 5 \times 1 \times \theta = 5\theta$.
Heat lost by water to reach temperature $\theta$ is $Q_3 = m_w c_w (40 - \theta) = 20 \times 1 \times (40 - \theta) = 800 - 20\theta$.
Equating heat gained and heat lost: $400 + 5\theta = 800 - 20\theta$.
$25\theta = 400$.
$\theta = \frac{400}{25} = 16\,^{\circ}C$.

(B) According to the principle of calorimetry,the total heat lost by the hotter liquids is equal to the total heat gained by the colder liquids.
Let $T$ be the final equilibrium temperature of the mixture.
The heat gained or lost by a substance is given by $Q = mc\Delta T$.
For the mixture,the total heat exchange must be zero:
$m_1 c_1 (T - T_1) + m_2 c_2 (T - T_2) + m_3 c_3 (T - T_3) = 0$
Expanding the terms:
$m_1 c_1 T - m_1 c_1 T_1 + m_2 c_2 T - m_2 c_2 T_2 + m_3 c_3 T - m_3 c_3 T_3 = 0$
Grouping the terms with $T$:
$T(m_1 c_1 + m_2 c_2 + m_3 c_3) = m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3$
Solving for $T$:
$T = \frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3}$

(D) According to the principle of calorimetry,heat lost by the aluminium sphere equals the heat gained by the water and the calorimeter.
Heat lost by aluminium: $Q_{lost} = m_{Al} \cdot S_{Al} \cdot \Delta T_{Al} = 0.20 \cdot S_{Al} \cdot (150 - 40) = 0.20 \cdot S_{Al} \cdot 110 = 22 \cdot S_{Al}$.
Heat gained by water: $Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T_{water} = 0.150 \cdot 4200 \cdot (40 - 27) = 0.150 \cdot 4200 \cdot 13 = 8190\, J$.
Heat gained by calorimeter: $Q_{cal} = W \cdot c_{water} \cdot \Delta T_{cal} = 0.025 \cdot 4200 \cdot (40 - 27) = 0.025 \cdot 4200 \cdot 13 = 1365\, J$.
Equating the heat: $22 \cdot S_{Al} = 8190 + 1365 = 9555$.
$S_{Al} = \frac{9555}{22} \approx 434.31\, J/kg\cdot ^\circ C$.
Rounding to the nearest integer,the specific heat of aluminium is $434\, J/kg\cdot ^\circ C$.

(A) Given:
Volume of water $V = 2\, L$,so mass $m = 2\, kg$ (since density of water is $1\, kg/L$).
Power of coil $P_{in} = 1000\, J/s$.
Rate of heat loss $P_{out} = 160\, J/s$.
Temperature change $\Delta T = 77\, ^\circ C - 27\, ^\circ C = 50\, ^\circ C$.
Specific heat capacity of water $c = 4.2\, kJ/kg\cdot K = 4200\, J/kg\cdot K$.
Net power supplied to the water $P_{net} = P_{in} - P_{out} = 1000\, J/s - 160\, J/s = 840\, J/s$.
Total heat required $Q = m \cdot c \cdot \Delta T = 2\, kg \times 4200\, J/kg\cdot K \times 50\, K = 420,000\, J$.
Time required $t = \frac{Q}{P_{net}} = \frac{420,000\, J}{840\, J/s} = 500\, s$.
Converting to minutes: $500\, s = 8\, min\, 20\, s$.

(B) According to the principle of calorimetry, $Heat \, lost \, by \, metal = Heat \, gained \, by \, calorimeter + Heat \, gained \, by \, water$.
Let $S$ be the specific heat of the metal in $J \, kg^{-1} \, K^{-1}$.
Mass of metal $m_m = 0.192 \, kg$, initial temperature $T_m = 100 \, ^oC$, final temperature $T_f = 21.5 \, ^oC$.
Mass of calorimeter $m_c = 0.128 \, kg$, specific heat $c_c = 394 \, J \, kg^{-1} \, K^{-1}$, initial temperature $T_i = 8.4 \, ^oC$.
Mass of water $m_w = 0.240 \, kg$, specific heat $c_w = 4186 \, J \, kg^{-1} \, K^{-1}$.
$0.192 \times S \times (100 - 21.5) = (0.128 \times 394 + 0.240 \times 4186) \times (21.5 - 8.4)$
$0.192 \times S \times 78.5 = (50.432 + 1004.64) \times 13.1$
$15.072 \times S = 1055.072 \times 13.1$
$15.072 \times S = 13821.4432$
$S \approx 917 \, J \, kg^{-1} \, K^{-1}$.
Rounding to the nearest provided option, the answer is $920 \, J \, kg^{-1} \, K^{-1}$.

(D) Let $m$ be the mass of ice added in grams.
Heat lost by water to reach $0\,^{\circ}C$: $Q_{lost} = m_w c_w \Delta T = 50 \times 4.2 \times (40 - 0) = 8400\,J.$
Heat gained by ice to reach $0\,^{\circ}C$: $Q_{gain,1} = m c_{ice} \Delta T = m \times 2.1 \times (0 - (-20)) = 42m\,J.$
Heat gained by the melted portion of ice to change phase at $0\,^{\circ}C$: $Q_{gain,2} = (m - 20) \times 334\,J.$
By the principle of calorimetry,$Q_{lost} = Q_{gain,1} + Q_{gain,2}.$
$8400 = 42m + 334(m - 20).$
$8400 = 42m + 334m - 6680.$
$8400 + 6680 = 376m.$
$15080 = 376m.$
$m = 15080 / 376 \approx 40.1\,g.$
Thus,the amount of ice added is approximately $40\,g.$

(D) Let the final equilibrium temperature be $T$.
According to the principle of calorimetry, heat lost by the metal ball = heat gained by the water and the vessel.
$m_{metal} c_{metal} (T_{initial, metal} - T) = (m_{water} c_{water} + C_{vessel}) (T - T_{initial, water})$
$0.1 \times 400 \times (500 - T) = (0.5 \times 4200 + 800) \times (T - 30)$
$40(500 - T) = (2100 + 800) \times (T - 30)$
$20000 - 40T = 2900(T - 30)$
$20000 - 40T = 2900T - 87000$
$107000 = 2940T$
$T = \frac{107000}{2940} \approx 36.39 \, ^\circ C$
Change in temperature $\Delta T = 36.39 - 30 = 6.39 \, ^\circ C$.
Percentage increment = $\frac{\Delta T}{T_{initial}} \times 100 = \frac{6.39}{30} \times 100 \approx 21.3 \%$.
Rounding to the nearest given option, the answer is $20 \%$.

(C) Using the principle of calorimetry,heat lost by liquid $A$ = heat gained by liquid $B$.
$m_A S_A (T_{A,initial} - T_{mix}) = m_B S_B (T_{mix} - T_{B,initial})$
$100 \times S_A \times (100 - 90) = 50 \times S_B \times (90 - 75)$
$1000 S_A = 750 S_B$
$S_A = 0.75 S_B = \frac{3}{4} S_B$
Now,for the second case,let the final temperature be $T$:
$100 \times S_A \times (100 - T) = 50 \times S_B \times (T - 50)$
Substitute $S_A = \frac{3}{4} S_B$:
$100 \times (\frac{3}{4} S_B) \times (100 - T) = 50 \times S_B \times (T - 50)$
$75 (100 - T) = 50 (T - 50)$
$3 (100 - T) = 2 (T - 50)$
$300 - 3T = 2T - 100$
$5T = 400$
$T = 80\,^oC$

(D) The total energy stored in the spring is converted into heat energy when the vibrations stop due to damping.
Energy stored in the spring $E = \frac{1}{2} k x^2$.
Given $k = 800\, N/m$ and $x = 2\, cm = 0.02\, m$.
$E = \frac{1}{2} \times 800 \times (0.02)^2 = 400 \times 0.0004 = 0.16\, J$.
This energy is absorbed by the mass and the water: $E = (m_1 s_1 + m_2 s_2) \Delta T$.
Here $m_1 = 0.5\, kg$,$s_1 = 400\, J/kg\, K$,$m_2 = 1\, kg$,$s_2 = 4184\, J/kg\, K$.
$0.16 = (0.5 \times 400 + 1 \times 4184) \Delta T$.
$0.16 = (200 + 4184) \Delta T = 4384 \Delta T$.
$\Delta T = \frac{0.16}{4384} \approx 3.65 \times 10^{-5}\, K$.
The order of magnitude is $10^{-5}\, K$.

(A) According to the principle of calorimetry,$Heat\,lost = Heat\,gained$.
Heat lost by $M_2$ grams of water cooling from $50\,^{\circ}C$ to $0\,^{\circ}C$ is given by $Q_{lost} = M_2 \times c_w \times \Delta T = M_2 \times 1 \times (50 - 0) = 50M_2$.
Heat gained by $M_1$ grams of ice warming from $-10\,^{\circ}C$ to $0\,^{\circ}C$ and then melting at $0\,^{\circ}C$ is given by $Q_{gained} = M_1 \times c_{ice} \times \Delta T + M_1 \times L_f = M_1 \times 0.5 \times 10 + M_1 \times L_f = 5M_1 + M_1 L_f$.
Equating the two: $50M_2 = 5M_1 + M_1 L_f$.
Rearranging for $L_f$: $M_1 L_f = 50M_2 - 5M_1$.
Therefore,$L_f = \frac{50M_2}{M_1} - 5$.

(A) Let the final equilibrium temperature be $T \, ^{\circ}C$.
Heat lost by boiling water = Heat gained by the brass vessel.
Heat lost by $20 \, g$ of water cooling from $100 \, ^{\circ}C$ to $T \, ^{\circ}C$ is $Q_{lost} = m_w c_w \Delta T_w = 20 \times 1 \times (100 - T)$.
Heat gained by $100 \, g$ of brass vessel warming from $0 \, ^{\circ}C$ to $T \, ^{\circ}C$ is $Q_{gained} = m_b c_b \Delta T_b = 100 \times 0.1 \times (T - 0)$.
Equating the two: $20(100 - T) = 10(T)$.
$2000 - 20T = 10T$.
$2000 = 30T$.
$T = \frac{2000}{30} = 66.66 \, ^{\circ}C$.

(C) Let the specific heat of ice be $S_{ice} = s$ and the specific heat of water be $S_{w} = 2s$.
Heat lost by $10\,g$ of water at $10\,^{\circ}C$ to reach $0\,^{\circ}C$ is:
$Q_{loss} = m_w S_w \Delta T = 10 \times (2s) \times (10 - 0) = 200s$.
Heat gained by $10\,g$ of ice at $-20\,^{\circ}C$ to reach $0\,^{\circ}C$ is:
$Q_{gain} = m_{ice} S_{ice} \Delta T = 10 \times s \times (0 - (-20)) = 200s$.
Since $Q_{loss} = Q_{gain}$,the heat released by the water is exactly enough to bring the ice to $0\,^{\circ}C$,and the water itself reaches $0\,^{\circ}C$. No phase change (melting or freezing) occurs because there is no excess heat available.
Therefore,at equilibrium,the system consists of $10\,g$ of ice and $10\,g$ of water at $0\,^{\circ}C$.

(A) Given:
Volume of water $V = 2\, L$,so mass $m = 2\, kg$.
Initial temperature $T_i = 27\, ^oC$,final temperature $T_f = 77\, ^oC$.
Change in temperature $\Delta T = 77 - 27 = 50\, ^oC$.
Power of heater $P = 1000\, W$.
Rate of heat loss $L = 160\, J/s$.
Specific heat of water $S = 4200\, J/(kg \cdot K)$.
The energy balance equation is:
$P \cdot t = L \cdot t + m \cdot S \cdot \Delta T$
$1000 \cdot t = 160 \cdot t + (2)(4200)(50)$
$840 \cdot t = 420000$
$t = \frac{420000}{840} = 500\, s$
Converting to minutes: $500\, s = 8\, min\ 20\, s$.

(B) The heat required to melt $60\,g$ of ice is given by $Q = mL$.
$Q = 60\,g \times 80\,cal/g = 4800\,cal$.
To convert this energy into Joules,we use the conversion factor $1\,cal = 4.2\,J$.
$Q = 4800 \times 4.2\,J = 20160\,J$.
The time taken is $1\,minute = 60\,seconds$.
Power $P = \frac{Q}{t} = \frac{20160\,J}{60\,s} = 336\,W$.

(B) Let $C$ be the thermal capacity of the calorimeter in $J/K$. The specific heat capacity of water is $s = 4200\, J/(kg\cdot K)$.
Heat lost by the hot water:
$Q_{lost} = m_{hot} \cdot s \cdot (T_{initial} - T_{final}) = 0.1 \times 4200 \times (60 - 35) = 0.1 \times 4200 \times 25 = 10500\, J$.
Heat gained by the cold water:
$Q_{cold} = m_{cold} \cdot s \cdot (T_{final} - T_{initial}) = 0.2 \times 4200 \times (35 - 30) = 0.2 \times 4200 \times 5 = 4200\, J$.
Heat gained by the calorimeter:
$Q_{cal} = C \cdot (T_{final} - T_{initial}) = C \times (35 - 30) = 5C$.
By the principle of calorimetry,Heat lost = Heat gained:
$10500 = 4200 + 5C$
$5C = 10500 - 4200$
$5C = 6300$
$C = 1260\, J/K$.

(C) Let $x$ be the mass of hot water required in $kg$.
According to the principle of calorimetry,heat lost by hot water equals heat gained by cold water.
$m_{hot} c (T_{hot} - T_{final}) = m_{cold} c (T_{final} - T_{cold})$
Here,$m_{hot} = x$,$T_{hot} = 100\,^{\circ}C$,$T_{final} = 35\,^{\circ}C$,$T_{cold} = 10\,^{\circ}C$,and $m_{cold} = (20 - x)$.
$x \cdot c \cdot (100 - 35) = (20 - x) \cdot c \cdot (35 - 10)$
$65x = (20 - x) \cdot 25$
$65x = 500 - 25x$
$90x = 500$
$x = \frac{500}{9} \approx 5.55\, kg$.
Rounding to the nearest provided option,the value is $5.6\, kg$.

(B) The principle of calorimetry states that heat lost by the hot body equals heat gained by the cold body.
Heat lost by the iron ball $= m_{iron} \cdot s_{iron} \cdot \Delta T$
Given: $m_{iron} = 0.2\,kg = 200\,g$,$\Delta T = (100\,^{\circ}C - 0\,^{\circ}C) = 100\,^{\circ}C$.
Heat gained by ice $= m_{ice} \cdot L_f$
Given: $m_{ice} = 25\,g$,$L_f = 80\,cal/g$.
Equating the two: $200 \cdot s_{iron} \cdot 100 = 25 \cdot 80$
$20000 \cdot s_{iron} = 2000$
$s_{iron} = \frac{2000}{20000} = 0.1\,cal/g\,^{\circ}C$.

(B) Let the final equilibrium temperature be $\theta$.
According to the principle of calorimetry,heat lost by the hot water equals the heat gained by the cold water and the beaker.
Heat lost by hot water $= m_h c (T_h - \theta) = 440 \times c \times (92 - \theta)$.
Heat gained by cold water $= m_c c (\theta - T_c) = 200 \times c \times (\theta - 20)$.
Heat gained by the beaker $= m_b c (\theta - T_c) = 20 \times c \times (\theta - 20)$.
Equating heat lost and heat gained:
$440(92 - \theta) = 200(\theta - 20) + 20(\theta - 20)$.
$440(92 - \theta) = 220(\theta - 20)$.
Dividing both sides by $220$:
$2(92 - \theta) = \theta - 20$.
$184 - 2\theta = \theta - 20$.
$3\theta = 204$.
$\theta = 68^{\circ}C$.

(B) According to the principle of calorimetry,heat lost by water equals heat gained by the solid ball.
Heat lost by water = $m \times S_w \times \Delta T_w = 200 \times S_w \times (80 - 60) = 200 \times S_w \times 20$.
Heat gained by solid = $m \times S_s \times \Delta T_s = 200 \times S_s \times (60 - 20) = 200 \times S_s \times 40$.
Equating the two: $200 \times S_w \times 20 = 200 \times S_s \times 40$.
$20 \times S_w = 40 \times S_s$.
$S_s = \frac{20}{40} \times S_w = 0.5 \times S_w$.
Therefore,the specific heat of the solid is one half of that of water.

(D) According to the principle of calorimetry,the heat lost by the water is equal to the heat gained by the ice.
Heat lost by water = $mc\Delta\theta = mc\theta$ (since the final temperature of water will be $0^oC$ as it reaches thermal equilibrium with ice).
Heat gained by ice = $m'L$,where $m'$ is the mass of ice melted.
Equating the two: $mc\theta = m'L$.
Therefore,the mass of ice melted is $m' = \frac{mc\theta}{L}$.

(A) Let $T \, ^oC$ be the final temperature of the mixture.
According to the principle of calorimetry,the heat lost by the hot water is equal to the heat gained by the cold water.
Heat lost by hot water = Heat gained by cold water
Mass $m = \text{density} (\rho) \times \text{volume} (V)$. Since the density of water is constant,we can write:
$V_1 \times \rho \times s \times (T_1 - T) = V_2 \times \rho \times s \times (T - T_2)$
Where $V_1 = 0.1 \, m^3$,$T_1 = 80 \, ^oC$,$V_2 = 0.3 \, m^3$,and $T_2 = 60 \, ^oC$.
$0.1 \times (80 - T) = 0.3 \times (T - 60)$
$8 - 0.1T = 0.3T - 18$
$26 = 0.4T$
$T = \frac{26}{0.4} = 65 \, ^oC$.

(D) Let $m$ be the mass of each liquid and $S_A, S_B, S_C$ be their specific heat capacities.
For the mixture of $A$ and $B$:
$m S_A (15 - 10) = m S_B (25 - 15)$
$5 S_A = 10 S_B \Rightarrow S_A = 2 S_B$
For the mixture of $B$ and $C$:
$m S_B (30 - 25) = m S_C (40 - 30)$
$5 S_B = 10 S_C \Rightarrow S_B = 2 S_C \Rightarrow S_C = \frac{S_B}{2}$
For the mixture of $A$ and $C$ with final temperature $\theta$:
$m S_A (\theta - 10) = m S_C (40 - \theta)$
Substitute $S_A = 2 S_B$ and $S_C = \frac{S_B}{2}$:
$2 S_B (\theta - 10) = \frac{S_B}{2} (40 - \theta)$
$4 (\theta - 10) = 40 - \theta$
$4\theta - 40 = 40 - \theta$
$5\theta = 80$
$\theta = 16^{\circ}C$

(B) The heat energy $Q$ required to raise the temperature of water is given by $Q = ms\Delta\theta$.
Here, mass $m = 1\, kg$ (since $1\, litre$ of water $\approx 1\, kg$), specific heat capacity $s = 4200\, J/kg\, ^oC$, and change in temperature $\Delta\theta = 60\, ^oC - 20\, ^oC = 40\, ^oC$.
$Q = 1\, kg \times 4200\, J/kg\, ^oC \times 40\, ^oC = 168,000\, J$.
The power of the heater $P = 1000\, W$.
The time $t$ required is given by $t = \frac{Q}{P}$.
$t = \frac{168,000\, J}{1000\, W} = 168\, seconds$.
Converting $168\, seconds$ into minutes and seconds: $168\, seconds = 2\, minutes$ and $48\, seconds$ ($2 \times 60 = 120$, $168 - 120 = 48$).

(A) Heat required to raise the temperature of $2\,kg$ of ice from $-20\,^oC$ to $0\,^oC$ is $Q_1 = m_i c_i \Delta T = 2000\,g \times 0.5\,cal/g^oC \times 20\,^oC = 20,000\,cal$.
Heat available from cooling $3\,kg$ of water from $15\,^oC$ to $0\,^oC$ is $Q_2 = m_w c_w \Delta T = 3000\,g \times 1\,cal/g^oC \times 15\,^oC = 45,000\,cal$.
Remaining heat available after warming the ice to $0\,^oC$ is $Q_{rem} = Q_2 - Q_1 = 45,000 - 20,000 = 25,000\,cal$.
Heat required to melt the entire $2\,kg$ of ice at $0\,^oC$ is $Q_3 = m_i L_f = 2000\,g \times 80\,cal/g = 160,000\,cal$.
Since $Q_{rem} < Q_3$,the available heat is not sufficient to melt all the ice. Therefore,the system will reach thermal equilibrium at $0\,^oC$ with a mixture of ice and water.

(B) The mass of $1\,L$ of water is $m = 1\,kg = 1000\,g$.
The specific heat capacity of water is $s = 4200\,J/(kg\cdot^{\circ}C)$.
The change in temperature is $\Delta \theta = 60\,^{\circ}C - 20\,^{\circ}C = 40\,^{\circ}C$.
The heat energy required is $Q = m s \Delta \theta = 1\,kg \times 4200\,J/(kg\cdot^{\circ}C) \times 40\,^{\circ}C = 168000\,J$.
The power of the heater is $P = 1000\,W$.
The time required is $t = Q / P = 168000\,J / 1000\,W = 168\,s$.
Converting $168\,s$ into minutes and seconds: $168\,s = 2\,min\, 48\,s$.

(A) Let $m$ be the mass of steam in grams. The density of water is $1\,g/cm^3$,so $6\,L = 6000\,g$.
Heat gained by water to reach $40\,^oC$: $Q_{gain} = m_{water} \cdot c_w \cdot \Delta T = 6000 \cdot 1 \cdot (40 - 20) = 120,000\,cal$.
Heat lost by steam at $100\,^oC$ to become water at $40\,^oC$: $Q_{lost} = m \cdot L_v + m \cdot c_w \cdot \Delta T = m \cdot 540 + m \cdot 1 \cdot (100 - 40) = 600m$.
Equating heat gained and lost: $120,000 = 600m$.
Solving for $m$: $m = \frac{120,000}{600} = 200\,g$.

(D) Let $m$ be the mass of steam condensed into water.
Heat lost by steam $=$ Heat gained by water.
Heat lost by steam consists of two parts: latent heat of condensation and cooling of condensed water from $100\,^{\circ}C$ to $80\,^{\circ}C$.
Heat lost $= m \times L_v + m \times s_w \times \Delta T_1 = m \times 540 + m \times 1 \times (100 - 80) = 540m + 20m = 560m$.
Heat gained by $20\,g$ of water to reach $80\,^{\circ}C$ from $10\,^{\circ}C$ is:
Heat gained $= m_w \times s_w \times \Delta T_2 = 20 \times 1 \times (80 - 10) = 20 \times 70 = 1400\,cal$.
Equating heat lost and gained: $560m = 1400$.
$m = \frac{1400}{560} = 2.5\,g$.
The total mass of water present is the initial mass plus the condensed steam mass:
Total mass $= 20\,g + 2.5\,g = 22.5\,g$.

(C) Mass of ice $(m_i)$ $= 1\,kg$,Initial temperature of ice $= 0\,^{\circ}C$.
Mass of water $(m_w)$ $= 1\,kg$,Initial temperature of water $= 80\,^{\circ}C$.
Latent heat of fusion of ice $(L_f)$ $= 336\,kJ/kg = 336,000\,J/kg$.
Specific heat of water $(c_w)$ $= 4200\,J/kg\cdot K$.
Heat required to melt $1\,kg$ of ice at $0\,^{\circ}C$ to water at $0\,^{\circ}C$ is $Q_{melt} = m_i \times L_f = 1 \times 336,000 = 336,000\,J$.
Heat released by $1\,kg$ of water cooling from $80\,^{\circ}C$ to $0\,^{\circ}C$ is $Q_{release} = m_w \times c_w \times \Delta T = 1 \times 4200 \times (80 - 0) = 336,000\,J$.
Since the heat released by the water is exactly equal to the heat required to melt the ice,the entire ice melts and the final temperature of the mixture remains $0\,^{\circ}C$.

(A) Total energy consumed by the kettle in $9.5$ minutes is given by $E = P \times t$.
$E = 2.5 \times 10^3 \, W \times (9.5 \times 60) \, s = 1425000 \, J$.
The heat required to raise the temperature of $3 \, kg$ of water from $15\, ^oC$ to $100\, ^oC$ is given by $Q = mc\Delta T$.
Using $c = 4200 \, J/(kg \cdot ^oC)$:
$Q = 3 \, kg \times 4200 \, J/(kg \cdot ^oC) \times (100 - 15)\, ^oC$.
$Q = 3 \times 4200 \times 85 = 1071000 \, J$.
The heat lost is the difference between the energy consumed and the heat absorbed by the water:
$\text{Heat lost} = E - Q = 1425000 \, J - 1071000 \, J = 354000 \, J$.
Rounding to the nearest provided option,the heat lost is $3.5 \times 10^5 \, J$.

(A) Let the final temperature be $T$.
Heat gained by ice to melt and reach temperature $T$ is given by $Q_{gain} = m_i L_f + m_i c_w (T - 0)$.
Substituting the values: $Q_{gain} = 10 \times 80 + 10 \times 1 \times T = 800 + 10T$.
Heat lost by the water and the tumbler (water equivalent $55\; g$) is given by $Q_{lost} = (m_w + W) c_w (40 - T)$.
Here,the water equivalent $W = 55\; g$ acts as additional mass of water. Assuming the initial mass of water in the tumbler is $55\; g$ (as implied by the water equivalent context),$Q_{lost} = 55 \times 1 \times (40 - T)$.
By the law of calorimetry,$Q_{gain} = Q_{lost}$.
$800 + 10T = 55(40 - T)$.
$800 + 10T = 2200 - 55T$.
$65T = 1400$.
$T = 1400 / 65 \approx 21.54^{\circ} C$.
Rounding to the nearest integer,$T \approx 22^{\circ} C$.