Principle of Calorimetry and Water Equivalent Questions in English

(C) Let $T$ be the resultant temperature of the mixture.
According to the principle of calorimetry,heat gained by the cold water equals the heat lost by the hot water.
Heat gained by $5 \ kg$ of water at $20^{\circ} C$ is:
$H_{\text{gain}} = m_1 c (T - T_1) = 5 \cdot c \cdot (T - 20) \quad \dots (i)$
Heat lost by $10 \ kg$ of water at $60^{\circ} C$ is:
$H_{\text{loss}} = m_2 c (T_2 - T) = 10 \cdot c \cdot (60 - T) \quad \dots (ii)$
Equating heat gained and heat lost:
$5c(T - 20) = 10c(60 - T)$
Dividing both sides by $5c$:
$T - 20 = 2(60 - T)$
$T - 20 = 120 - 2T$
$3T = 140$
$T = \frac{140}{3} \approx 46.67^{\circ} C$
Rounding to the nearest whole number,the resultant temperature is $47^{\circ} C$.

(B) Let the mass of the steam condensed be $x \, kg$.
Heat lost by steam = Heat gained by water + Heat gained by calorimeter.
Heat lost by steam = (Heat released during condensation) + (Heat released by condensed water cooling from $100^{\circ} C$ to $90^{\circ} C$).
$Q_{lost} = x \cdot L + x \cdot c_w \cdot \Delta T_1 = x \cdot 540 + x \cdot 1 \cdot (100 - 90) = 540x + 10x = 550x \, kcal$.
Heat gained by water = $m_w \cdot c_w \cdot \Delta T_2 = 1 \, kg \cdot 1 \, kcal \cdot kg^{-1} \cdot {}^{\circ} C^{-1} \cdot (90 - 9)^{\circ} C = 81 \, kcal$.
Heat gained by calorimeter = $W \cdot c_w \cdot \Delta T_2 = 0.1 \, kg \cdot 1 \, kcal \cdot kg^{-1} \cdot {}^{\circ} C^{-1} \cdot (90 - 9)^{\circ} C = 8.1 \, kcal$.
Equating heat lost and gained: $550x = 81 + 8.1 = 89.1$.
$x = \frac{89.1}{550} = 0.162 \, kg = 162 \, g$.

(B) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dQ}{dt} = -k(T_{avg} - T_0)$.
Since $dQ = (m + x)c \Delta T$,where $x$ is the water equivalent of the calorimeter,we have $(m + x)c \frac{\Delta T}{t} = k(T_{avg} - T_0)$.
For the first case: $(75 + x)c \frac{(62 - 58)}{9} = k(T_{avg} - T_0) \implies (75 + x) \frac{4}{9} = K'$ (where $K'$ is a constant).
For the second case: $(105 + x)c \frac{(62 - 58)}{12} = k(T_{avg} - T_0) \implies (105 + x) \frac{4}{12} = K'$.
Equating the two expressions: $\frac{75 + x}{9} = \frac{105 + x}{12}$.
Multiplying both sides by $36$: $4(75 + x) = 3(105 + x)$.
$300 + 4x = 315 + 3x$.
$x = 315 - 300 = 15 \ g$.

(B) Given: Mass of ice $m_i = 2 \ kg$,Temperature of ice $T_i = -20^{\circ} C$,Mass of water $m_w = 5 \ kg$,Temperature of water $T_w = 20^{\circ} C$.
Specific heat of ice $c_i = 2100 \ J/(kg \cdot K)$,Specific heat of water $c_w = 4200 \ J/(kg \cdot K)$,Latent heat of fusion $L_f = 3.36 \times 10^5 \ J/kg$.
Step $1$: Heat required to bring ice to $0^{\circ} C$: $Q_1 = m_i c_i \Delta T = 2 \times 2100 \times 20 = 84,000 \ J$.
Step $2$: Heat released by water to reach $0^{\circ} C$: $Q_2 = m_w c_w \Delta T = 5 \times 4200 \times 20 = 420,000 \ J$.
Step $3$: Heat remaining after warming ice to $0^{\circ} C$: $Q_{rem} = Q_2 - Q_1 = 420,000 - 84,000 = 336,000 \ J$.
Step $4$: Heat required to melt all ice: $Q_3 = m_i L_f = 2 \times 3.36 \times 10^5 = 672,000 \ J$.
Since $Q_{rem} < Q_3$,only part of the ice melts. Mass of ice melted $m_{melt} = Q_{rem} / L_f = 336,000 / 336,000 = 1 \ kg$.
Total mass of water = Initial water + Melted ice = $5 \ kg + 1 \ kg = 6 \ kg$.

(C) Let the mass of each liquid be $m$. The principle of calorimetry states that the heat lost equals the heat gained,or more generally,the sum of $ms\Delta T$ for all components is zero at equilibrium temperature $T_f$.
Sum of heat capacities: $\sum m_i s_i T_i = \sum m_i s_i T_f$.
Since $m$ is constant,$m \sum_{n=1}^{10} (ns)(10n) = m \sum_{n=1}^{10} (ns) T_f$.
Dividing by $ms$,we get $\sum_{n=1}^{10} 10n^2 = T_f \sum_{n=1}^{10} n$.
Using summation formulas: $\sum_{n=1}^{10} n^2 = \frac{10(11)(21)}{6} = 385$ and $\sum_{n=1}^{10} n = \frac{10(11)}{2} = 55$.
Substituting these values: $10(385) = T_f(55)$.
$3850 = 55 T_f$.
$T_f = \frac{3850}{55} = 70^{\circ} C$.

(B) For equal masses $m$,the final temperature $T_{mix}$ of a mixture of two liquids with specific heats $C_1, C_2$ and initial temperatures $T_1, T_2$ is given by $T_{mix} = \frac{T_1 C_1 + T_2 C_2}{C_1 + C_2}$.
For liquids $A$ and $B$:
$20 = \frac{15 C_A + 24 C_B}{C_A + C_B} \Rightarrow 20 C_A + 20 C_B = 15 C_A + 24 C_B \Rightarrow 5 C_A = 4 C_B \Rightarrow C_A = \frac{4}{5} C_B$.
For liquids $B$ and $C$:
$26 = \frac{24 C_B + 30 C_C}{C_B + C_C} \Rightarrow 26 C_B + 26 C_C = 24 C_B + 30 C_C \Rightarrow 2 C_B = 4 C_C \Rightarrow C_B = 2 C_C$.
Expressing all in terms of $C_C$:
$C_B = 2 C_C$
$C_A = \frac{4}{5} (2 C_C) = \frac{8}{5} C_C$
Thus,$C_A : C_B : C_C = \frac{8}{5} C_C : 2 C_C : C_C = 8 : 10 : 5$.

(C) By the principle of calorimetry,Heat lost by water $=$ Heat gained by ice.
Let $T$ be the final equilibrium temperature.
Heat lost by $74 \ g$ of water $= m_w c_w (T_i - T) = 74 \times 1 \times (70 - T)$.
Heat gained by $37 \ g$ of ice to melt at $0^{\circ} C = m_i L_f = 37 \times 80$.
Heat gained by $37 \ g$ of melted water to reach temperature $T = m_i c_w (T - 0) = 37 \times 1 \times T$.
Equating the heat lost and gained:
$74(70 - T) = 37 \times 80 + 37T$.
Dividing by $37$:
$2(70 - T) = 80 + T$.
$140 - 2T = 80 + T$.
$3T = 60$.
$T = 20^{\circ} C$.

(A) For water,$m_1 = 360 \ g$,$T_1 = 40^{\circ} C$,$T_f = 100^{\circ} C$.
Heat required by water to reach $100^{\circ} C$ is $H_1 = m_1 C_w (T_f - T_1) = 360 \times 1 \times (100 - 40) = 21600 \ cal$.
At equilibrium,heat released by steam condensing at $100^{\circ} C$ equals the heat required by water.
Let $m'$ be the mass of steam that condenses. $m' \times L = H_1 \implies m' \times 540 = 21600 \implies m' = 40 \ g$.
Remaining mass of steam $m_s = 60 - 40 = 20 \ g$.
Total mass of water $m_w = 360 + 40 = 400 \ g$.
The ratio of masses of steam to water is $m_s : m_w = 20 : 400 = 1 : 20$.

(C) According to the principle of calorimetry,Heat lost by hot body = Heat gained by cold body.
Let the final temperature be $T$.
Heat lost by $74 \ g$ of water at $70^{\circ} C$ to reach temperature $T$ is $Q_{lost} = m_w c_w (70 - T) = 74 \times 1 \times (70 - T)$.
Heat gained by $37 \ g$ of ice at $0^{\circ} C$ to melt and reach temperature $T$ is $Q_{gained} = m_i L_f + m_i c_w (T - 0) = 37 \times 80 + 37 \times 1 \times T$.
Equating the two: $74(70 - T) = 2960 + 37T$.
$5180 - 74T = 2960 + 37T$.
$111T = 2220$.
$T = 20^{\circ} C$.

(D) According to the principle of calorimetry, the heat lost by the steam equals the heat gained by the water.
Heat gained by water: $Q_{gain} = m_w c_w \Delta T_w = 100 \,g \times 1 \,cal/g^{\circ}C \times (90^{\circ}C - 24^{\circ}C) = 100 \times 66 = 6600 \,cal$.
Heat lost by steam: $Q_{lost} = m_s L_v + m_s c_w \Delta T_s = m_s(540) + m_s(1)(100^{\circ}C - 90^{\circ}C) = 540 m_s + 10 m_s = 550 m_s$.
Equating the two: $550 m_s = 6600$.
$m_s = \frac{6600}{550} = 12 \,g$.

(A) According to the principle of calorimetry, Heat gained by water $=$ Heat lost by steam.
Let $m$ be the mass of steam added.
Heat gained by water: $Q_1 = m_w s_w \Delta T_w = 150 \times 1 \times (40 - 20) = 150 \times 20 = 3000 \,cal$.
Heat lost by steam: $Q_2 = m L_v + m s_w \Delta T_s = m \times 540 + m \times 1 \times (100 - 40) = m(540 + 60) = 600m \,cal$.
Equating the two: $3000 = 600m$.
$m = \frac{3000}{600} = 5 \,g$.
The total mass of water at $40^{\circ} C$ is the initial mass of water plus the mass of condensed steam: $150 \,g + 5 \,g = 155 \,g$.

(C) First,calculate the mass of water: $m = V \times \rho = 3 \times 10^{-3} \text{ m}^3 \times 1000 \text{ kg/m}^3 = 3 \text{ kg}$.
Heat required to raise the temperature: $\Delta Q = m S \Delta T = 3 \times 4200 \times (80 - 0) = 1,008,000 \text{ J}$.
Power of the heater: $P = \frac{V^2}{R} = \frac{200^2}{50} = \frac{40000}{50} = 800 \text{ W}$.
Time required: $t = \frac{\Delta Q}{P} = \frac{1,008,000}{800} = 1260 \text{ s}$.
Converting to minutes: $t = \frac{1260}{60} = 21 \text{ min}$.

(C) Given: Power of heater $P = 210 \,W$, mass of water $m = 100 \,g = 0.1 \,kg$, specific heat capacity $c = 4200 \,J / kg \cdot ^{\circ} C$, initial temperature $T_1 = 25^{\circ} C$, and final temperature $T_2 = 100^{\circ} C$.
Change in temperature $\Delta T = T_2 - T_1 = 100^{\circ} C - 25^{\circ} C = 75^{\circ} C$.
The heat energy $Q$ required to raise the temperature is given by $Q = m c \Delta T$.
Substituting the values: $Q = 0.1 \,kg \times 4200 \,J / kg \cdot ^{\circ} C \times 75^{\circ} C = 31500 \,J$.
Since $P = Q / t$, the time $t$ required is $t = Q / P$.
$t = 31500 \,J / 210 \,W = 150 \,s$.
Therefore, the time required is $150 \,s$.

(A) According to the principle of calorimetry, Heat lost by water = Heat gained by ice.
Let $m_i$ be the mass of ice added.
Heat lost by water: $Q_1 = m_w c_w \Delta T_w = (0.2 \,kg) \times (4190 \,J/kg^{\circ} C) \times (20^{\circ} C - 0^{\circ} C) = 16760 \,J$.
Heat gained by ice to reach $0^{\circ} C$: $Q_2 = m_i c_i \Delta T_i = m_i \times (2100 \,J/kg^{\circ} C) \times (0^{\circ} C - (-10^{\circ} C)) = 21000 m_i \,J$.
Heat gained by ice to melt at $0^{\circ} C$: $Q_3 = m_i L_F = m_i \times (3.34 \times 10^5 \,J/kg) = 334000 m_i \,J$.
Equating heat lost and gained: $16760 = 21000 m_i + 334000 m_i$.
$16760 = 355000 m_i$.
$m_i = \frac{16760}{355000} \approx 0.0472 \,kg = 47.2 \,g$.
Rounding to the nearest integer, the mass of ice required is $47 \,g$.

(C) Let the final temperature at equilibrium be $T$.
According to the principle of calorimetry,the heat lost by the coffee equals the heat gained by the milk.
Heat lost by coffee = $m_c \cdot c_c \cdot (T_i - T)$
Heat gained by milk = $m_m \cdot c_m \cdot (T - T_m)$
Given: $m_c = 250 \ g$,$c_c = 1.00 \ cal/g^{\circ} C$,$T_i = 90^{\circ} C$,$m_m = 20 \ g$,$c_m = 1.00 \ cal/g^{\circ} C$,$T_m = 5^{\circ} C$.
Equating the two:
$250 \times 1.00 \times (90 - T) = 20 \times 1.00 \times (T - 5)$
$250(90 - T) = 20(T - 5)$
$22500 - 250T = 20T - 100$
$22600 = 270T$
$T = \frac{22600}{270} \approx 83.7^{\circ} C$

(B) According to the principle of calorimetry:
Heat lost by steam = Heat gained by water and calorimeter
Let $m$ be the mass of steam condensed in grams.
Heat lost by steam = $m \times L + m \times C_w \times (T_{steam} - T_{final})$
$= m \times 540 + m \times 1 \times (100 - 90) = 550m \ cal$
Heat gained by water and calorimeter = $(m_{water} + m_{eq}) \times C_w \times (T_{final} - T_{initial})$
$= (1000 \ g + 200 \ g) \times 1 \times (90 - 9) = 1200 \times 81 = 97200 \ cal$
Equating the two:
$550m = 97200$
$m = \frac{97200}{550} \approx 176.7 \ g$
Converting to $kg$,$m \approx 0.1767 \ kg$,which is approximately $0.18 \ kg$.

(B) Heat released by $300 \text{ g}$ of water when cooling from $25^{\circ}C$ to $0^{\circ}C$ is given by $Q_{released} = m \cdot c \cdot \Delta T$.
$Q_{released} = 300 \text{ g} \times 1 \text{ cal/g}^{\circ}C \times (25^{\circ}C - 0^{\circ}C) = 7500 \text{ cal}$.
Heat required to melt $100 \text{ g}$ of ice at $0^{\circ}C$ into water at $0^{\circ}C$ is given by $Q_{required} = m \cdot L_f$.
$Q_{required} = 100 \text{ g} \times 80 \text{ cal/g} = 8000 \text{ cal}$.
Since $Q_{required} > Q_{released}$,the available heat is insufficient to melt all the ice.
Therefore,the mixture will reach a state of thermal equilibrium at $0^{\circ}C$ with some ice remaining unmelted.
The final temperature of the mixture is $0^{\circ}C$.

(D) Let $W$ be the water equivalent of the calorimeter in $g$.
Step $1$: Heat lost by hot water = Heat gained by cold water + Heat gained by calorimeter.
$100 \times 1 \times (100 - 20) = 300 \times 1 \times (20 - 10) + W \times 1 \times (20 - 10)$
$8000 = 3000 + 10W$
$10W = 5000 \implies W = 500 \ g$.
Step $2$: Heat lost by mixture = Heat gained by metallic block.
Mass of mixture = $100 \ g + 300 \ g = 400 \ g$.
Heat lost by mixture and calorimeter = $(400 \times 1 + 500) \times (20 - 19) = 900 \times 1 = 900 \ cal$.
Heat gained by metallic block = $m \times S_b \times \Delta T = 1000 \ g \times S_b \times (19 - 10) = 9000 \times S_b$.
Equating the two: $9000 \times S_b = 900$.
$S_b = 0.1 \ cal/g^{\circ} C$.

(C) Let the initial mass of ice be $m \ g$.
According to the principle of calorimetry,heat lost by the system (calorimeter + water) = heat gained by the ice.
The system cools down to $0^{\circ} C$ because only half of the ice melts,implying the final equilibrium temperature is $0^{\circ} C$.
Heat lost by water and calorimeter: $Q_{lost} = (m_{water} + m_{cal}) \cdot s_{water} \cdot \Delta T = (50 + 10) \cdot 1.0 \cdot (15 - 0) = 60 \cdot 15 = 900 \ cal$.
Heat gained by ice: The ice warms from $-10^{\circ} C$ to $0^{\circ} C$ and then half of it melts.
$Q_{gained} = m \cdot s_{ice} \cdot \Delta T_{ice} + (m/2) \cdot L_f = m \cdot 0.5 \cdot (0 - (-10)) + (m/2) \cdot 80$.
$Q_{gained} = m \cdot 0.5 \cdot 10 + 40m = 5m + 40m = 45m$.
Equating heat lost and gained: $900 = 45m$.
$m = 900 / 45 = 20 \ g$.

(B) Let the final equilibrium temperature be $\theta$. According to the principle of calorimetry,the heat lost by the hotter bodies must equal the heat gained by the colder bodies.
Heat lost by the body at $60^{\circ} C$ (mass $3m$) = $3m \cdot s \cdot (60 - \theta)$
Heat gained by the body at $50^{\circ} C$ (mass $m$) = $m \cdot s \cdot (\theta - 50)$
Heat gained by the body at $40^{\circ} C$ (mass $m$) = $m \cdot s \cdot (\theta - 40)$
Equating heat lost and heat gained:
$3ms(60 - \theta) = ms(\theta - 50) + ms(\theta - 40)$
Dividing by $ms$:
$3(60 - \theta) = (\theta - 50) + (\theta - 40)$
$180 - 3\theta = 2\theta - 90$
$5\theta = 270$
$\theta = 54^{\circ} C$

(B) The heat supplied by the heater is absorbed by both the liquid and the container. The formula is: $(m_L s_L + m_C s_C) \Delta T = P \times t$.
For water: $m_w = 0.5 \ kg$,$s_w = 4200 \ J \ kg^{-1} \ K^{-1}$,$\Delta T = 3 \ K$,$t_1 = 15 \times 60 = 900 \ s$,$P = 10 \ W$.
$(0.5 \times 4200 + m_C s_C) \times 3 = 10 \times 900$
$(2100 + m_C s_C) \times 3 = 9000$
$2100 + m_C s_C = 3000 \implies m_C s_C = 900 \ J \ K^{-1}$.
For oil: $m_o = 2 \ kg$,$s_o = ?$,$\Delta T = 2 \ K$,$t_2 = 20 \times 60 = 1200 \ s$,$P = 10 \ W$.
$(2 \times s_o + m_C s_C) \times 2 = 10 \times 1200$
$(2 s_o + 900) \times 2 = 12000$
$2 s_o + 900 = 6000$
$2 s_o = 5100$
$s_o = 2550 \ J \ kg^{-1} \ K^{-1} = 2.55 \times 10^{3} \ J \ kg^{-1} \ K^{-1}$.

(C) Step $1$: Calculate heat required to bring $5 \ g$ of ice from $-20^{\circ} C$ to $0^{\circ} C$: $Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T = 5 \ g \times 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1} \times 20^{\circ} C = 50 \ cal$.
Step $2$: Calculate heat required to melt $5 \ g$ of ice at $0^{\circ} C$: $Q_2 = m_{ice} \cdot L_f = 5 \ g \times 80 \ cal \ g^{-1} = 400 \ cal$.
Total heat required to convert ice to water at $0^{\circ} C$ is $Q_{total} = 50 + 400 = 450 \ cal$.
Step $3$: Calculate heat released by $19 \ g$ of water cooling from $30^{\circ} C$ to $0^{\circ} C$: $Q_{released} = m_{water} \cdot c_{water} \cdot \Delta T = 19 \ g \times 1 \ cal \ g^{-1} (^{\circ} C)^{-1} \times 30^{\circ} C = 570 \ cal$.
Since $Q_{released} > Q_{total}$,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.
$19 \times 1 \times (30 - T_f) = 450 + 5 \times 1 \times (T_f - 0)$.
$570 - 19 T_f = 450 + 5 T_f$.
$120 = 24 T_f$.
$T_f = 5^{\circ} C$.

(A) Heat gained by ice = Heat lost by water.
Heat gained by ice = (Heat to raise ice from $-10^{\circ}C$ to $0^{\circ}C$) + (Heat to melt ice at $0^{\circ}C$) + (Heat to raise melted water from $0^{\circ}C$ to $T^{\circ}C$).
$Q_{gain} = (10 \times 2100 \times 10) + (10 \times 3.36 \times 10^5) + (10 \times 4200 \times T) = 210000 + 3360000 + 42000T = 3570000 + 42000T$.
Heat lost by water = $100 \times 4200 \times (25 - T) = 420000 \times (25 - T) = 10500000 - 420000T$.
Equating the two: $3570000 + 42000T = 10500000 - 420000T$.
$462000T = 6930000$.
$T = 6930000 / 462000 = 15^{\circ}C$.
The decrement in temperature is $\Delta T = 25^{\circ}C - 15^{\circ}C = 10^{\circ}C$.