Mix Examples-Thermometry, Thermal Expansion and Calorimetry Questions in English

(D) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation: $E = \alpha t + \frac{1}{2} \beta t^2$,where $t$ is the temperature difference between the junctions.
This equation represents a parabola.
As the temperature difference $t$ increases,the $e.m.f.$ $(E)$ first increases,reaches a maximum value,and then decreases.
Therefore,the graph that represents this variation is a parabolic curve that opens downwards,which corresponds to the shape shown in Graph $D$.

(A) The thermoelectric electromotive force $(E)$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the hot and cold junctions.
The thermoelectric power $(P)$ is defined as the rate of change of thermoelectric $EMF$ with respect to the temperature difference:
$P = \frac{dE}{d\theta} = \frac{d}{d\theta} (\alpha \theta + \frac{1}{2} \beta \theta^2) = \alpha + \beta \theta$.
Comparing this equation with the linear equation $y = mx + c$,where $y = P$,$x = \theta$,$m = \beta$ (which is negative for a thermocouple),and $c = \alpha$,we see that the thermoelectric power varies linearly with the temperature difference $\theta$.
Therefore,the graph representing this linear relationship is a straight line with a negative slope,which corresponds to option $A$.

(D) The thermo $e.m.f.$ is given by $E = \alpha\theta + \frac{1}{2}\beta\theta^2$.
Comparing the given equation $E = 40\theta - \frac{\theta^2}{20}$ with the standard equation,we get $\alpha = 40$ and $\frac{1}{2}\beta = -\frac{1}{20}$,which implies $\beta = -\frac{1}{10}$.
The neutral temperature $\theta_n$ is the temperature at which the thermo $e.m.f.$ is maximum,given by the condition $\frac{dE}{d\theta} = 0$.
$\frac{d}{d\theta}(40\theta - \frac{\theta^2}{20}) = 40 - \frac{2\theta}{20} = 40 - \frac{\theta}{10} = 0$.
Solving for $\theta$,we get $\theta = 40 \times 10 = 400\,^oC$.

(A) The change in volume due to thermal expansion is given by $\Delta V_{thermal} = V \alpha \Delta \theta$,where $\Delta \theta$ is the change in temperature.
The change in volume due to pressure $P$ is given by the definition of bulk modulus: $\beta = -\frac{P}{\Delta V / V}$,which implies $\Delta V_{pressure} = -\frac{PV}{\beta}$.
For the cube to regain its original volume,the net change in volume must be zero: $\Delta V_{thermal} + \Delta V_{pressure} = 0$.
$V \alpha \Delta \theta - \frac{PV}{\beta} = 0$.
$V \alpha \Delta \theta = \frac{PV}{\beta}$.
$\Delta \theta = \frac{P}{\alpha \beta}$.

(A) The total heat $Q$ required is the sum of heat for four stages:
$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \cdot 0.5 \cdot 10 = 5\,cal$.
$2$. Melting ice at $0^{\circ}C$: $Q_2 = m \cdot L_{f} = 1 \cdot 80 = 80\,cal$.
$3$. Heating water from $0^{\circ}C$ to $100^{\circ}C$: $Q_3 = m \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot 100 = 100\,cal$.
$4$. Vaporizing water at $100^{\circ}C$: $Q_4 = m \cdot L_{v} = 1 \cdot 540 = 540\,cal$.
Total heat $Q = 5 + 80 + 100 + 540 = 725\,cal$.
Converting to Joules: $Q = 725 \cdot 4.2 = 3045\,J$.

(D) The thermo $EMF$ in a thermocouple is given by the equation $E = a\theta + \frac{1}{2}b\theta^2$.
Comparing this with the given equation $E = 40\theta - \frac{1}{20}\theta^2$,we get $a = 40$ and $b = -\frac{1}{10}$.
The neutral temperature $\theta_n$ is the temperature at which the $EMF$ is maximum,which occurs when $\frac{dE}{d\theta} = 0$.
$\frac{dE}{d\theta} = 40 - \frac{2\theta}{20} = 40 - \frac{\theta}{10}$.
Setting $\frac{dE}{d\theta} = 0$,we get $40 - \frac{\theta}{10} = 0$.
Therefore,$\theta = 400\;^oC$.

(B) The total thermoelectric $emf$ generated between $0 \, ^oC$ and $100 \, ^oC$ is the sum of the $emf$ generated in the intermediate temperature intervals.
Given:
$e_0^{100} = 200 \, \mu V$
$e_0^{32} = 64 \, \mu V$
$e_{32}^{70} = 76 \, \mu V$
Using the additive property of thermoelectric $emf$:
$e_0^{100} = e_0^{32} + e_{32}^{70} + e_{70}^{100}$
Substituting the given values:
$200 = 64 + 76 + e_{70}^{100}$
$200 = 140 + e_{70}^{100}$
$e_{70}^{100} = 200 - 140 = 60 \, \mu V$
Therefore,the $emf$ generated between $70 \, ^oC$ and $100 \, ^oC$ is $60 \, \mu V$.

(D) The thermo-emf $(E)$ produced by a single thermocouple is given by $E = \alpha \Delta T$,where $\alpha$ is the Seebeck coefficient and $\Delta T$ is the temperature difference.
Given $\alpha = 40 \, \mu V/K$.
The temperature difference $\Delta T = |40 \, ^\circ C - 20 \, ^\circ C| = 20 \, ^\circ C = 20 \, K$.
Thus,$E = 40 \, \mu V/K \times 20 \, K = 800 \, \mu V$.
For $N = 150$ thermocouples connected in series,the total emf $(E_{total})$ is $E_{total} = N \times E$.
$E_{total} = 150 \times 800 \, \mu V = 120,000 \, \mu V$.
Since $1 \, mV = 1000 \, \mu V$,we have $E_{total} = 120 \, mV$.

(A) When a substance is heated at a constant rate,its temperature increases until it reaches a phase transition point (melting or boiling). During the phase transition,the temperature remains constant as the heat supplied is used to change the state of the substance (latent heat).
For oxygen,the melting point is approximately $54 \, K$ and the boiling point is approximately $90 \, K$.
As the substance is heated from $50 \, K$ to $300 \, K$,it will undergo phase changes (melting and boiling).
$1$. Initially,the temperature of solid/liquid oxygen increases linearly with time.
$2$. At the melting point,the temperature remains constant while the solid melts.
$3$. Then,the temperature of the liquid increases until it reaches the boiling point.
$4$. At the boiling point,the temperature remains constant while the liquid boils into gas.
$5$. Finally,the temperature of the gas increases.
The graph that shows these plateaus (constant temperature regions) during phase changes is the correct representation. Among the given options,the graph showing a linear increase,a plateau,and another linear increase corresponds to the heating curve of a substance undergoing phase changes.

(B) The potential energy of the ice at height $h$ is $PE = mgh$.
According to the problem, all this energy is converted into heat during the fall.
Only one-quarter of this heat is absorbed by the ice to melt it completely.
The heat required to melt a mass $m$ of ice is $Q = mL$, where $L$ is the latent heat of fusion.
Therefore, the energy balance equation is $\frac{1}{4} (mgh) = mL$.
Canceling $m$ from both sides, we get $\frac{gh}{4} = L$.
Solving for $h$, we get $h = \frac{4L}{g}$.
Substituting the given values: $h = \frac{4 \times 3.4 \times 10^{5}}{10} \text{ m}$.
$h = 4 \times 3.4 \times 10^{4} \text{ m} = 13.6 \times 10^{4} \text{ m} = 136,000 \text{ m}$.
Converting to kilometers, $h = 136 \text{ km}$.

(B) When salt crystals dissolve,the crystal lattice is destroyed. This process requires a certain amount of energy (latent heat of dissolution),which is absorbed from the water,causing the temperature to drop.
In vessel $B$,the salt is in powder form,meaning a significant part of the intermolecular bonds has already been destroyed during the crushing process. Consequently,less energy is required to dissolve the powder compared to the large crystals in vessel $A$. Since less energy is absorbed from the water in vessel $B$,the final temperature of the solution in vessel $B$ will be higher than in vessel $A$.

(D) The heat released by $x \, g$ of steam at $100^\circ C$ condensing into water at $100^\circ C$ is given by $Q_{released} = x \times L_v$,where $L_v = 540 \, cal/g$.
So,$Q_{released} = 540x \, cal$.
The heat required to convert $y \, g$ of ice at $0^\circ C$ into water at $100^\circ C$ involves two steps: melting the ice and heating the resulting water.
$Q_{absorbed} = (y \times L_f) + (y \times c_w \times \Delta T)$,where $L_f = 80 \, cal/g$,$c_w = 1 \, cal/g^\circ C$,and $\Delta T = 100^\circ C$.
$Q_{absorbed} = 80y + 100y = 180y \, cal$.
Equating the heat released and absorbed: $540x = 180y$.
Therefore,$\frac{y}{x} = \frac{540}{180} = \frac{3}{1}$.
The ratio $y:x$ is $3:1$.

(B) Let the initial volume of the liquid be $V_0 = A_0 l_0$,where $A_0$ is the cross-sectional area and $l_0$ is the length of the liquid column.
When the system is heated by a temperature change $\Delta T$,the new volume of the liquid is $V_L = V_0(1 + \gamma \Delta T)$.
The new cross-sectional area of the glass tube is $A = A_0(1 + 2\alpha \Delta T)$,where $2\alpha$ is the coefficient of area expansion.
Since the length of the liquid column remains constant at $l_0$,the new volume of the liquid must be equal to the new volume of the tube up to height $l_0$,which is $V_{tube} = A \times l_0 = A_0(1 + 2\alpha \Delta T)l_0 = V_0(1 + 2\alpha \Delta T)$.
Equating the two volumes: $V_0(1 + \gamma \Delta T) = V_0(1 + 2\alpha \Delta T)$.
This simplifies to $1 + \gamma \Delta T = 1 + 2\alpha \Delta T$,which gives $\gamma = 2\alpha$.

(C) The potential energy of the water at the top is converted into heat energy at the bottom.
$mgh = mc\Delta \theta$
Here,$m$ is the mass of water,$g = 9.8 \ m/s^2$,$h = 500 \ m$,and $c = 4200 \ J/kg \cdot ^\circ C$.
Canceling $m$ from both sides,we get $\Delta \theta = \frac{gh}{c}$.
Substituting the values: $\Delta \theta = \frac{9.8 \times 500}{4200} = \frac{4900}{4200} = \frac{49}{42} = \frac{7}{6} \approx 1.166 \ ^\circ C$.
Rounding to the nearest option,the rise in temperature is $1.16 \ ^\circ C$.

(B) According to the law of conservation of energy,the loss in potential energy of the ball is converted into heat energy,which is absorbed by the ball,causing a rise in its temperature.
The loss in potential energy is given by:
$\Delta PE = mg(h_1 - h_2)$
This energy is equal to the heat absorbed by the ball:
$Q = mc\Delta \theta$
Equating the two:
$mg(h_1 - h_2) = mc\Delta \theta$
Solving for the rise in temperature $\Delta \theta$:
$\Delta \theta = \frac{g(h_1 - h_2)}{c}$
Given:
$g = 10 \, m \, s^{-2}$
$h_1 = 10 \, m$
$h_2 = 5.4 \, m$
$c = 460 \, J \, kg^{-1} \, ^\circ C^{-1}$
$\Delta \theta = \frac{10 \times (10 - 5.4)}{460}$
$\Delta \theta = \frac{10 \times 4.6}{460}$
$\Delta \theta = \frac{46}{460} = 0.1 \, ^\circ C$

(C) The process of converting ice at $0^\circ C$ to water at $100^\circ C$ occurs in two steps:
Step $1$: Melting of ice at $0^\circ C$ to water at $0^\circ C$.
The heat required is $Q_1 = m \cdot L_f$,where $m = 1 \, g$ and $L_f = 80 \, cal/g$.
$Q_1 = 1 \times 80 = 80 \, cal$.
Step $2$: Heating water from $0^\circ C$ to $100^\circ C$.
The heat required is $Q_2 = m \cdot c \cdot \Delta T$,where $m = 1 \, g$,$c = 1 \, cal/g^\circ C$,and $\Delta T = (100 - 0) = 100^\circ C$.
$Q_2 = 1 \times 1 \times 100 = 100 \, cal$.
Total heat required $Q = Q_1 + Q_2 = 80 + 100 = 180 \, cal$.

(B) The heat supplied is given by $\Delta Q = mc\Delta T$.
Substituting the values: $20000 = 1 \times 400 \times \Delta T$.
Thus,$\Delta T = \frac{20000}{400} = 50\,^oC$.
The work done at constant atmospheric pressure is $W = P\Delta V$.
Since $\Delta V = V_0 \gamma \Delta T$,we have $W = P V_0 \gamma \Delta T$.
The initial volume $V_0 = \frac{m}{\rho} = \frac{1}{9000}\,m^3$.
Substituting the values: $W = (10^5) \times (\frac{1}{9000}) \times (9 \times 10^{-5}) \times 50$.
$W = 10^5 \times \frac{1}{9 \times 10^3} \times 9 \times 10^{-5} \times 50 = 10^5 \times 10^{-8} \times 50 = 10^{-3} \times 50 = 0.05\,J$.

(D) The temperature difference $\Delta T = T_{hot} - T_{cold} = 40\,^{\circ}C - 20\,^{\circ}C = 20\,^{\circ}C = 20\,K$.
The emf developed by a single thermocouple is $E_1 = \alpha \times \Delta T = 40\,\mu V/K \times 20\,K = 800\,\mu V$.
Since $150$ thermocouples are connected in series,the total emf $E_{total} = n \times E_1 = 150 \times 800\,\mu V$.
$E_{total} = 120,000\,\mu V = 120\,mV$.

(C) The coefficient of volume expansion of the vessel is $\gamma_{\text{vessel}} = 3 \alpha_{\text{vessel}} = 3 \times 10^{-3} \ ^\circ C^{-1}$.
The coefficient of volume expansion of the liquid is $\gamma_{\text{liquid}} = 10^{-3} \ ^\circ C^{-1}$.
Since $\gamma_{\text{vessel}} > \gamma_{\text{liquid}}$,the volume of the vessel increases more than the volume of the liquid upon heating. This causes the liquid level to fall,resulting in a decrease in pressure at the bottom.
The change in volume of the vessel relative to the liquid is given by $\Delta V_{\text{rel}} = V(\gamma_{\text{vessel}} - \gamma_{\text{liquid}}) \Delta T$.
Substituting the values: $\Delta V_{\text{rel}} = V(3 \times 10^{-3} - 1 \times 10^{-3}) \times 10 = V(2 \times 10^{-3}) \times 10 = 0.02 V$.
Since the cross-sectional area $A$ of the cylinder also increases,the change in height $\Delta h$ is related to the change in volume. For a thin-walled vessel,the fractional change in height is $\frac{\Delta h}{h} \approx \frac{\Delta V}{V} = 0.02$.
Pressure $P = h \rho g$. The fractional change in pressure is $\frac{\Delta P}{P} = \frac{\Delta h}{h} = 0.02$.
Therefore,the pressure decreases by $0.02 \times 100\% = 2\%$.

(C) Let $T$ be the equilibrium temperature.
At equilibrium,the diameter of the copper ring equals the diameter of the aluminium sphere:
$D_{Cu}(T) = D_{Al}(T) \implies 25(1 + \alpha_{Cu}T) = 25.05(1 + \alpha_{Al}(T - 100))$.
Substituting values: $25(1 + 17 \times 10^{-6}T) = 25.05(1 + 2.3 \times 10^{-5}(T - 100))$.
Solving for $T$: $25 + 4.25 \times 10^{-4}T = 25.05 + 5.7615 \times 10^{-4}(T - 100)$.
$25 + 4.25 \times 10^{-4}T = 25.05 + 5.7615 \times 10^{-4}T - 0.057615$.
$0.007615 = 1.5115 \times 10^{-4}T \implies T \approx 50.38^o C$.
By calorimetry: $m_{Cu} s_{Cu} (T - 0) = m_{Al} s_{Al} (100 - T)$.
$m_{Cu} (0.0923)(50.38) = m_{Al} (0.215)(100 - 50.38)$.
$m_{Cu} (4.65) = m_{Al} (10.67)$.
Ratio $m_{Al} / m_{Cu} = 4.65 / 10.67 \approx 0.435 \approx 23/54$.

(C) The apparent weight of a body immersed in a liquid is given by $W_{app} = W_{actual} - F_B$,where $F_B$ is the buoyant force.
The buoyant force is given by $F_B = V_{sub} \cdot \rho_{liquid} \cdot g$,where $V_{sub}$ is the volume of the submerged body and $\rho_{liquid}$ is the density of the liquid.
As the temperature increases,both the volume of the metal ball and the density of the alcohol change. The volume of the metal ball increases as $V_m(T) = V_0(1 + \gamma_m \Delta T)$ and the density of alcohol decreases as $\rho_{Al}(T) = \frac{\rho_0}{1 + \gamma_{Al} \Delta T}$.
The buoyant force at temperature $T$ is $F_B(T) = V_m(T) \cdot \rho_{Al}(T) \cdot g = V_0(1 + \gamma_m \Delta T) \cdot \frac{\rho_0}{1 + \gamma_{Al} \Delta T} \cdot g$.
Since it is given that $\gamma_{Al} > \gamma_m$,the denominator $(1 + \gamma_{Al} \Delta T)$ increases more than the numerator $(1 + \gamma_m \Delta T)$. Therefore,the buoyant force $F_B$ decreases as the temperature increases.
Since $W_{app} = W_{actual} - F_B$ and $F_B$ decreases,the apparent weight $W_{app}$ must increase. Thus,$W_2 > W_1$ or $W_1 < W_2$.

(B) Total heat supplied in $4$ minutes $(240 \, s)$ is $Q = 100 \, cal/s \times 240 \, s = 24000 \, cal$.
Step $1$: Heat required to raise the temperature of the system from $-20^{\circ} C$ to $0^{\circ} C$:
$Q_1 = m_{Al} c_{Al} \Delta T + m_{ice} c_{ice} \Delta T = (100 \times 0.2 \times 20) + (200 \times 0.5 \times 20) = 400 + 2000 = 2400 \, cal$.
Time taken for this step $= 2400 / 100 = 24 \, s$.
Step $2$: Heat required to melt the ice at $0^{\circ} C$:
$Q_2 = m_{ice} L = 200 \times 80 = 16000 \, cal$.
Time taken for this step $= 16000 / 100 = 160 \, s$.
Total time elapsed $= 24 + 160 = 184 \, s$. Remaining time $= 240 - 184 = 56 \, s$.
Remaining heat available $= 24000 - (2400 + 16000) = 5600 \, cal$.
Step $3$: Heat used to raise the temperature of the system (water + container) from $0^{\circ} C$ to $\theta$:
$Q_3 = (m_{Al} c_{Al} + m_{water} c_{water}) \Delta \theta = (100 \times 0.2 + 200 \times 1) \theta = (20 + 200) \theta = 220 \theta$.
$220 \theta = 5600 \implies \theta = 5600 / 220 \approx 25.45^{\circ} C \approx 25.5^{\circ} C$.

(A) Let $m_1$ be the mass of water that freezes and $m_2$ be the mass of water that vaporizes.
Since the vessel is thermally insulated, the heat released by the freezing water is equal to the heat absorbed by the vaporizing water.
$m_1 L_f = m_2 L_v$
Given $L_f = 3.36 \times 10^5 \text{ J/kg}$ and $L_v = 21 \times 10^5 \text{ J/kg}$.
$m_1 (3.36 \times 10^5) = m_2 (21 \times 10^5)$
$m_2 = \frac{3.36}{21} m_1 = 0.16 m_1$
The total initial mass of water is $M = m_1 + m_2 = m_1 + 0.16 m_1 = 1.16 m_1$.
The percentage of water solidified is $\frac{m_1}{M} \times 100\% = \frac{m_1}{1.16 m_1} \times 100\%$.
$= \frac{100}{1.16} \% \approx 86.2 \%$.

(D) The total heat required consists of the heat to melt the ice,heat to raise the temperature of the resulting water and the calorimeter to $100^{\circ} C$,and the heat to vaporize the water.
$1$. Heat to melt $10 \, g$ of ice: $Q_1 = m L_f = 10 \, g \times 80 \, cal/g = 800 \, cal$.
$2$. Heat to raise the temperature of $10 \, g$ of water (from ice) and $10 \, g$ of water equivalent (calorimeter) from $0^{\circ} C$ to $100^{\circ} C$: $Q_2 = (m_{water} + m_{eq}) C \Delta T = (10 + 10) \times 1 \times 100 = 2000 \, cal$.
$3$. Heat to vaporize $10 \, g$ of water at $100^{\circ} C$: $Q_3 = m L_v = 10 \, g \times 540 \, cal/g = 5400 \, cal$.
Total heat $Q = Q_1 + Q_2 + Q_3 = 800 + 2000 + 5400 = 8200 \, cal$.

(A) The mass of ice melted in $100 \, s$ is $m = 0.1 \, g/s \times 100 \, s = 10 \, g$.
The heat supplied to melt the ice is $Q = m L_f$,where $L_f = 80 \, cal/g$ (latent heat of fusion of ice).
$Q = 10 \, g \times 80 \, cal/g = 800 \, cal$.
The rate of heat supply is $P = Q / t = 800 \, cal / 100 \, s = 8 \, cal/s$.
After the ice melts,we have $10 \, g$ of water. The rate of rise of temperature is given by $P = m_{water} c_{water} (dT/dt)$.
$8 \, cal/s = 10 \, g \times 1 \, cal/(g \cdot ^\circ C) \times (dT/dt)$.
$(dT/dt) = 8 / 10 = 0.8 \, ^\circ C/s$.

(B) Given: Power rating $P = 2000 \, W$,Efficiency $\eta = 0.8$,Flow rate $v = 100 \, cc/s$,Inlet temperature $T_i = 10^{\circ} C$.
Since $1 \, cc$ of water has a mass of $1 \, g$,the mass flow rate is $m = 100 \, g/s = 0.1 \, kg/s$.
The effective heat power supplied to the water is $Q = P \times \eta = 2000 \times 0.8 = 1600 \, J/s$.
The formula for heat transfer is $Q = m \cdot c \cdot \Delta T$,where $c$ is the specific heat capacity of water $(4200 \, J/kg \cdot ^{\circ} C)$.
Substituting the values: $1600 = 0.1 \times 4200 \times (T_o - 10)$.
$1600 = 420 \times (T_o - 10)$.
$T_o - 10 = 1600 / 420 \approx 3.81$.
$T_o = 13.81^{\circ} C$. Thus,the outlet temperature is approximately $13.8^{\circ} C$.

(C) Let $x$ be the mass of ice that melts into water.
When ice melts into water,the volume decreases because the density of water is higher than that of ice.
The volume of $x \, g$ of ice is $V_{ice} = \frac{x}{\rho_{ice}} = \frac{x}{0.9 \rho_{water}}$.
The volume of $x \, g$ of water is $V_{water} = \frac{x}{\rho_{water}}$.
The change in volume is $\Delta V = V_{ice} - V_{water} = \frac{x}{0.9 \rho_{water}} - \frac{x}{\rho_{water}} = 1 \, cm^3$.
Assuming $\rho_{water} = 1 \, g/cm^3$,we have $\frac{x}{0.9} - x = 1$.
$x(1.111 - 1) = 1 \Rightarrow x(0.1/0.9) = 1 \Rightarrow x = 9 \, g$.
The heat required to melt $9 \, g$ of ice is $Q = m \times L = 9 \, g \times 80 \, cal/g = 720 \, cal$.

(B) The rate of heat absorption is proportional to the surface area $A$ of the sphere.
$Q = mL$,where $m$ is the mass and $L$ is the latent heat of fusion.
Rate of heat absorption $\frac{dQ}{dt} = L \frac{dm}{dt} = L \rho \frac{dV}{dt}$,where $\rho$ is the density and $V$ is the volume.
Since $V = \frac{4}{3} \pi r^3$,we have $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given $\frac{dQ}{dt} \propto A$,where $A = 4 \pi r^2$,we have $L \rho (4 \pi r^2 \frac{dr}{dt}) = -k (4 \pi r^2)$,where $k$ is a constant.
Simplifying this,we get $\frac{dr}{dt} = -\frac{k}{L \rho} = -C$,where $C$ is a positive constant.
Integrating this,$r(t) = R - Ct$.
This represents a linear decrease in radius with time,which is depicted by graph $B$.

(D) The process involves two stages:
$1$. Heating the ice from $-5^oC$ to $0^oC$: $Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T$.
Given $m$ is in grams, $m_{ice} = m \times 10^{-3} \ kg$.
$Q_1 = (m \times 10^{-3}) \times 2100 \times 5 = 10.5m \ J$.
$2$. Melting $1 \ gm$ of ice at $0^oC$: $Q_2 = m_{melted} \cdot L_f$.
$Q_2 = (1 \times 10^{-3}) \times 3.36 \times 10^5 = 336 \ J$.
Total heat given $Q = Q_1 + Q_2 = 420 \ J$.
$10.5m + 336 = 420$.
$10.5m = 84$.
$m = 84 / 10.5 = 8 \ gm$.

(C) Let the constant rate of heat supply be $P$ $(J/s)$.
Heat required for melting (fusion) is $Q_f = P \times (90 - 10) = 80P$. Since $Q_f = mL_f$,we have $mL_f = 80P$.
Heat required for vaporization is $Q_v = P \times (120 - 100) = 20P$. Since $Q_v = mL_v$,we have $mL_v = 20P$.
Comparing the two,$mL_f = 80P$ and $mL_v = 20P$,so $L_f > L_v$.
For the heating phases,the slope of the temperature-time graph is given by $\frac{dT}{dt} = \frac{P}{mS}$. Thus,$S = \frac{P}{m(dT/dt)}$.
The slope for the solid phase is $m_s = \frac{30 - 10}{10} = 2 \, ^\circ C/s$.
The slope for the liquid phase is $m_l = \frac{200 - 30}{100 - 90} = \frac{170}{10} = 17 \, ^\circ C/s$.
Since $m_l > m_s$,and $S \propto \frac{1}{\text{slope}}$,it follows that $S_s > S_l$,or $S_l < S_s$.
Therefore,$L_f > L_v$ and $S_l < S_s$.

(A) Heat required to raise the temperature of $2 \ kg$ ice from $-20^{\circ} C$ to $0^{\circ} C$ is $Q_1 = m_i c_i \Delta T = 2000 \ g \times 0.5 \ cal/g^{\circ} C \times 20^{\circ} C = 20000 \ cal$.
Heat released by $5 \ kg$ water cooling from $20^{\circ} C$ to $0^{\circ} C$ is $Q_2 = m_w c_w \Delta T = 5000 \ g \times 1 \ cal/g^{\circ} C \times 20^{\circ} C = 100000 \ cal$.
Remaining heat available to melt the ice is $Q_{rem} = Q_2 - Q_1 = 100000 - 20000 = 80000 \ cal$.
Mass of ice that melts is $m = Q_{rem} / L_f = 80000 \ cal / 80 \ cal/g = 1000 \ g = 1 \ kg$.
Final amount of water $= \text{Initial water} + \text{Melted ice} = 5 \ kg + 1 \ kg = 6 \ kg$.

(D) Let the cross-sectional area of the beaker be $A$. The initial volume of the liquid is $V_0 = A H$.
When heated by temperature $T$,the new volume of the liquid is $V_L = V_0(1 + \gamma_L T) = AH(1 + \alpha T)$,where $\gamma_L = \alpha$ is the volume expansion coefficient of the liquid.
The new cross-sectional area of the beaker is $A' = A(1 + 2\alpha_s T)$,where $\alpha_s = 3\alpha$ is the linear expansion coefficient of the beaker material. Thus,$A' = A(1 + 2(3\alpha)T) = A(1 + 6\alpha T)$.
The new height of the beaker is $H' = H(1 + \alpha_s T) = H(1 + 3\alpha T)$.
The new height of the liquid $H_L$ in the beaker is $H_L = V_L / A' = AH(1 + \alpha T) / [A(1 + 6\alpha T)]$.
Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $H_L \approx H(1 + \alpha T)(1 - 6\alpha T) \approx H(1 + \alpha T - 6\alpha T) = H(1 - 5\alpha T)$.
The reduction in the level of the liquid is $\Delta H = H' - H_L = H(1 + 3\alpha T) - H(1 - 5\alpha T) = H(1 + 3\alpha T - 1 + 5\alpha T) = 8\alpha TH$.

(A) Let $m$ be the total mass of the material, $L$ be the latent heat of fusion, and $s$ be the specific heat of the liquid.
In Process-$1$, the energy $Q$ melts $\frac{2}{3}$ of the mass:
$Q = \frac{2}{3} m L$ --- (Equation $1$)
In Process-$2$, an additional energy $Q$ is added. This energy first melts the remaining $\frac{1}{3}$ of the solid and then raises the temperature of the entire liquid mass $m$ from $0^\circ C$ to $40^\circ C$:
$Q = \frac{1}{3} m L + m s (40 - 0)$
$Q = \frac{1}{3} m L + 40 m s$ --- (Equation $2$)
Equating Equation $1$ and Equation $2$:
$\frac{2}{3} m L = \frac{1}{3} m L + 40 m s$
$\frac{1}{3} m L = 40 m s$
$\frac{L}{s} = 40 \times 3 = 120$
Thus, the ratio of the latent heat of fusion to the specific heat of the liquid is $120$.

(D) Let the equilibrium temperature be $T$. Since the system is adiabatic,the total heat lost by the iron rod equals the total heat gained by the air and the vessel-piston assembly.
Heat lost by the iron rod = $C(8T_0 - T)$.
Heat gained by the vessel and piston = $2C(T - T_0)$.
Heat gained by $2$ moles of diatomic gas (at constant pressure,as the piston is movable and atmospheric pressure $P_0$ is constant) = $n C_p (T - T_0) = 2 \times \frac{7R}{2} (T - T_0) = 7R(T - T_0)$.
Equating heat lost and gained: $C(8T_0 - T) = 2C(T - T_0) + 7R(T - T_0)$.
$8CT_0 - CT = 2CT - 2CT_0 + 7RT - 7RT_0$.
$8CT_0 + 2CT_0 + 7RT_0 = CT + 2CT + 7RT$.
$T_0(10C + 7R) = T(3C + 7R)$.
$T = \left( \frac{10C + 7R}{3C + 7R} \right) T_0$.

(A) Let the mass of the bullet be $m\ g$.
The total heat required for the bullet to just melt is given by $Q_1 = mc\Delta T + mL$.
Substituting the values: $Q_1 = m \times 0.03 \times (327 - 27) + m \times 6 = m \times 0.03 \times 300 + 6m = 9m + 6m = 15m\ cal$.
Converting this to Joules: $Q_1 = 15m \times 4.2 = 63m\ J$.
When the bullet is stopped,the loss in its kinetic energy is $\Delta K = \frac{1}{2} m_{kg} v^2 = \frac{1}{2} (m \times 10^{-3}) v^2\ J$.
Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet.
Energy absorbed by the bullet $Q_2 = 0.75 \times \frac{1}{2} \times m \times 10^{-3} \times v^2 = \frac{3}{8} \times 10^{-3} \times m v^2\ J$.
For the bullet to just melt,$Q_2 = Q_1$.
$\frac{3}{8} \times 10^{-3} \times m v^2 = 63m$.
$v^2 = \frac{63 \times 8 \times 1000}{3} = 21 \times 8000 = 168000$.
$v = \sqrt{168000} \approx 409.87\ m/s \approx 410\ m/s$.

(A) Step $1$: Calculate heat required to melt $1 \ gm$ of ice at $0^o C$ to water at $0^o C$: $Q_1 = m \cdot L_f = 1 \ gm \times 80 \ cal/gm = 80 \ cal$.
Step $2$: Calculate heat released by $1 \ gm$ of steam at $100^o C$ condensing to water at $100^o C$: $Q_2 = m \cdot L_v = 1 \ gm \times 540 \ cal/gm = 540 \ cal$.
Step $3$: Since $Q_2 > Q_1$,the ice melts completely and the resulting water at $0^o C$ is heated by the remaining heat from the steam.
Step $4$: Heat available to raise the temperature of the water: $Q_{rem} = Q_2 - Q_1 = 540 - 80 = 460 \ cal$.
Step $5$: Total mass of water is $2 \ gm$. Let the final temperature be $T$. The heat absorbed by $2 \ gm$ of water to reach $T$ is $Q = m \cdot c \cdot \Delta T = 2 \ gm \times 1 \ cal/gm^o C \times (T - 0) = 2T$.
Step $6$: Equating heat gained and lost: $2T = 460$,which gives $T = 230^o C$. However,since the steam is at $100^o C$,the final temperature cannot exceed $100^o C$. This implies all steam condenses and the mixture reaches equilibrium at $100^o C$ with some steam remaining.

(B) The process involves two steps:
$1$. Melting $5\ kg$ of ice at $0\ ^oC$ to water at $0\ ^oC$. The latent heat of fusion of ice is $L_f = 80\ kcal/kg$.
Heat required $Q_1 = m \times L_f = 5\ kg \times 80\ kcal/kg = 400\ kcal$.
$2$. Heating $5\ kg$ of water from $0\ ^oC$ to $100\ ^oC$. The specific heat capacity of water is $c = 1\ kcal/(kg \cdot ^oC)$.
Heat required $Q_2 = m \times c \times \Delta T = 5\ kg \times 1\ kcal/(kg \cdot ^oC) \times (100\ ^oC - 0\ ^oC) = 500\ kcal$.
Total heat required $Q = Q_1 + Q_2 = 400\ kcal + 500\ kcal = 900\ kcal$.

(B) Heat required to melt $1 \ kg$ of ice into water at $0^{\circ} C = 80 \ kcal$.
Heat required to raise the temperature of $1 \ kg$ of water from $0^{\circ} C$ to $100^{\circ} C = mS \Delta t = (1 \ kg) \times (1 \ kcal/kg^{\circ} C) \times (100 - 0) = 100 \ kcal$.
Heat required to convert $1 \ kg$ of boiling water into steam at $100^{\circ} C = mL = 1 \ kg \times 540 \ kcal/kg = 540 \ kcal$.
Thus,the total heat required to convert $1 \ kg$ of ice at $0^{\circ} C$ into steam at $100^{\circ} C = 80 + 100 + 540 = 720 \ kcal$.

(B) The initial volume of the flask and mercury is $V_0 = 10^3 \ cc$.
The change in temperature is $\Delta T = 100 \, ^oC - 0 \, ^oC = 100 \, ^oC$.
The increase in the volume of the glass flask is given by $\Delta V_g = V_0 \gamma_g \Delta T$.
$\Delta V_g = 10^3 \times 40 \times 10^{-6} \times 100 = 4 \, cc$.
The increase in the volume of the mercury is given by $\Delta V_m = V_0 \gamma_m \Delta T$.
$\Delta V_m = 10^3 \times 180 \times 10^{-6} \times 100 = 18 \, cc$.
The volume of mercury that overflows is the difference between the expansion of mercury and the expansion of the flask:
$\Delta V_{overflow} = \Delta V_m - \Delta V_g = 18 \, cc - 4 \, cc = 14 \, cc$.

(C) The density of water is maximum at $4^{\circ}C$,which implies that the volume of a given mass of water is minimum at this temperature.
When the temperature increases above $4^{\circ}C$,water expands,causing the volume to increase,which leads to overflow.
When the temperature decreases below $4^{\circ}C$,water also expands (anomalous expansion of water),causing the volume to increase,which again leads to overflow.
Therefore,water overflows on both heating and cooling from $4^{\circ}C$.

(A) The process involves four stages:
$1$. Heating ice from $-10\,^{\circ}C$ to $0\,^{\circ}C$: $Q_1 = m \cdot s_{ice} \cdot \Delta T = 1 \times 0.5 \times 10 = 5\,cal$.
$2$. Melting ice at $0\,^{\circ}C$ to water at $0\,^{\circ}C$: $Q_2 = m \cdot L_f = 1 \times 80 = 80\,cal$.
$3$. Heating water from $0\,^{\circ}C$ to $100\,^{\circ}C$: $Q_3 = m \cdot s_{water} \cdot \Delta T = 1 \times 1 \times 100 = 100\,cal$.
$4$. Converting water at $100\,^{\circ}C$ to steam at $100\,^{\circ}C$: $Q_4 = m \cdot L_v = 1 \times 540 = 540\,cal$.
Total heat $Q = Q_1 + Q_2 + Q_3 + Q_4 = 5 + 80 + 100 + 540 = 725\,cal$.
Converting to Joules: $Q = 725 \times 4.2 = 3045\,J$.

(B) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$. The change in time period due to temperature change is given by $\Delta t = \frac{1}{2} \alpha \Delta \theta t$,where $\alpha$ is the coefficient of linear expansion,$\Delta \theta$ is the change in temperature,and $t$ is the total time.
Let $\theta$ be the temperature at which the clock gives the correct time.
At $15 \ ^oC$,the clock is $5 \ s$ fast,so $\Delta t_1 = 5 = \frac{1}{2} \alpha (\theta - 15) \times (24 \times 3600) \quad ...(1)$
At $30 \ ^oC$,the clock is $10 \ s$ slow,so $\Delta t_2 = 10 = \frac{1}{2} \alpha (30 - \theta) \times (24 \times 3600) \quad ...(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{5}{10} = \frac{\theta - 15}{30 - \theta}$
$\frac{1}{2} = \frac{\theta - 15}{30 - \theta}$
$30 - \theta = 2(\theta - 15)$
$30 - \theta = 2\theta - 30$
$3\theta = 60$
$\theta = 20 \ ^oC$.

(B) When the balls are heated,they expand,and their radii increase. This causes the center of mass of each ball to shift.
For ball $A$,which is suspended from the ceiling,the expansion causes its center of mass to move downward. As the center of mass moves downward,the gravitational potential energy of ball $A$ decreases. This decrease in potential energy is converted into internal energy,providing additional heating to the ball.
For ball $B$,which rests on the floor,the expansion causes its center of mass to move upward. As the center of mass moves upward,the gravitational potential energy of ball $B$ increases. This increase in potential energy requires energy,which is taken from the absorbed heat,effectively reducing the energy available for increasing the temperature of the ball.
Since ball $A$ gains additional internal energy from its gravitational potential energy change while ball $B$ loses some of its absorbed heat to increase its gravitational potential energy,the final temperature of ball $A$ will be higher than that of ball $B$. Therefore,$T_A > T_B$.

(A) When the surface of a lake freezes,the ice acts as an insulator.
Since the density of water is maximum at $4\,^oC$,the water at the bottom of the lake remains at $4\,^oC$ due to convection currents that stop once the water reaches this temperature.
The layer of water in direct contact with the lower surface of the ice is in thermal equilibrium with the ice,which is at $0\,^oC$.
Therefore,the temperature gradient exists from $0\,^oC$ at the ice-water interface to $4\,^oC$ at the bottom of the lake.

(D) The total heat removed $(Q)$ consists of three parts:
$1$. Heat to cool water from $25\,^{\circ}C$ to $0\,^{\circ}C$: $Q_1 = m \cdot c_w \cdot \Delta T = 500 \times 1 \times 25 = 12500\,cal$.
$2$. Heat to freeze water at $0\,^{\circ}C$ to ice at $0\,^{\circ}C$: $Q_2 = m \cdot L_f = 500 \times 80 = 40000\,cal$.
$3$. Heat to cool ice from $0\,^{\circ}C$ to $-10\,^{\circ}C$: $Q_3 = m \cdot c_i \cdot \Delta T = 500 \times 0.5 \times 10 = 2500\,cal$.
Total heat $Q = Q_1 + Q_2 + Q_3 = 12500 + 40000 + 2500 = 55000\,cal$.
Total time $t = 3\,hours\,40\,minutes = (3 \times 60) + 40 = 220\,minutes$.
Heat removed per minute = $\frac{Q}{t} = \frac{55000}{220} = 250\,cal/min$.

(B) bimetallic strip consists of two different metals joined together,which have different coefficients of linear expansion.
When the temperature changes,the two metals expand or contract by different amounts due to the principle of thermal expansion $(iv)$.
This difference in expansion causes the strip to bend,which is a form of energy conversion $(ii)$ where thermal energy is converted into mechanical work (bending).
Therefore,the functioning of a bimetallic strip depends on both energy conversion and thermal expansion.

(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 released by $3\, kg$ of water cooling from $15\,^oC$ to $0\,^oC$ is $Q_2 = m_w c_w \Delta T = 3000 \, g \times 1 \, cal/g^oC \times 15^oC = 45,000 \, cal$.
Net heat available to melt the ice at $0\,^oC$ is $Q_{net} = Q_2 - Q_1 = 45,000 - 20,000 = 25,000 \, cal$.
Heat required to melt all the ice is $Q_{melt} = m_i L_f = 2000 \, g \times 80 \, cal/g = 160,000 \, cal$.
Since $Q_{net} < Q_{melt}$,the ice will not melt completely. The system will reach an equilibrium state at $0\,^oC$ with a mixture of ice and water.

(B) Let $V_g$ be the volume of the glass container and $V_l$ be the volume of the liquid at temperature $T$. Let $\gamma_g$ and $\gamma_l$ be the coefficients of volume expansion of the glass and liquid,respectively.
Given $\gamma_g : \gamma_l = 1 : 4$,so $\gamma_l = 4\gamma_g$.
The vacant space is $V_v = V_g - V_l$.
For the vacant space to remain constant at all temperatures,the change in volume of the container must equal the change in volume of the liquid.
$\Delta V_g = \Delta V_l$
$V_g \gamma_g \Delta T = V_l \gamma_l \Delta T$
$V_g \gamma_g = V_l (4\gamma_g)$
$V_g = 4V_l$
Therefore,the fraction of the volume occupied by the liquid is $\frac{V_l}{V_g} = \frac{1}{4}$.

(B) $1$. Heat released by $10\, g$ of steam at $100\,^{\circ}C$ to become water at $100\,^{\circ}C$: $Q_1 = m_s L_v = 10 \times 540 = 5400\, cal$.
$2$. Heat released by $10\, g$ of hot water at $100\,^{\circ}C$ to reach $0\,^{\circ}C$: $Q_2 = m_s c_w \Delta T = 10 \times 1 \times 100 = 1000\, cal$.
$3$. Total heat available to melt ice: $Q_{total} = 5400 + 1000 = 6400\, cal$.
$4$. Heat required to melt $100\, g$ of ice at $0\,^{\circ}C$: $Q_{ice} = m_i L_f = 100 \times 80 = 8000\, cal$.
$5$. Since $Q_{total} < Q_{ice}$,only a portion of ice melts. Mass of ice melted: $m_{melted} = Q_{total} / L_f = 6400 / 80 = 80\, g$.
$6$. Total water in the calorimeter = (Initial water) + (Mass of ice melted) + (Mass of condensed steam) = $500 + 80 + 10 = 590\, g$.

(D) Let $m$ be the mass of water that evaporates in grams.
Since the process is adiabatic and no external heat is supplied,the heat released by the freezing of a portion of the water must be equal to the heat absorbed by the portion that evaporates.
Mass of water that turns into ice $= (150 - m)\, g$.
Heat released during freezing $\Delta Q_{rel} = (150 - m) \times L_f$.
Heat absorbed during evaporation $\Delta Q_{req} = m \times L_v$.
Equating the two: $(150 - m) \times L_f = m \times L_v$.
Substituting the values $L_f = 3.36 \times 10^5\, J/kg$ and $L_v = 2.10 \times 10^6\, J/kg$:
$(150 - m) \times 3.36 \times 10^5 = m \times 21.0 \times 10^5$.
$150 \times 3.36 = m \times (21.0 + 3.36)$.
$504 = m \times 24.36$.
$m = 504 / 24.36 \approx 20.69\, g$.
The mass of evaporated water is closest to $20\, g$.