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

(C) The total heat $Q$ required to raise the temperature of $1, kg$ of water from $20, ^oC$ to $100, ^oC$ and then to evaporate it is given by:
$Q = mc\Delta T + mL$
Here, $m = 1, kg$, $c = 4200, J/kg, ^oC$, $\Delta T = (100 - 20) = 80, ^oC$, and $L = 2260 \times 10^3, J/kg$.
$Q = (1 \times 4200 \times 80) + (1 \times 2260 \times 10^3) = 336000 + 2260000 = 2596000, J$.
The power $P$ dissipated by the heating element is:
$P = \frac{V_{rms}^2}{R} = \frac{200^2}{20} = \frac{40000}{20} = 2000, W$.
The time $t$ taken is given by $t = \frac{Q}{P}$:
$t = \frac{2596000}{2000} = 1298, s$.
Converting to minutes: $t = \frac{1298}{60} \approx 21.63, min \approx 22, min$.

(B) The temperature of the body is given by the function $T = 6t^2 + 4$.
To find the rate of change of temperature with respect to time,we differentiate $T$ with respect to $t$:
$\frac{dT}{dt} = \frac{d}{dt}(6t^2 + 4) = 12t$.
Now,we evaluate this rate at $t = 10 \ s$:
$\frac{dT}{dt} \Big|_{t=10} = 12 \times 10 = 120 \ K/s$.
Therefore,the rate of change of temperature at $t = 10 \ s$ is $120 \ K/s$.

(A) Power of the drilling machine $P = 10 \, kW = 10^4 \, W$.
Time $t = 2.5 \, minutes = 2.5 \times 60 = 150 \, s$.
Total energy supplied $E = P \times t = 10^4 \times 150 = 1.5 \times 10^6 \, J$.
Since $50 \%$ of the power is lost,the energy absorbed by the aluminium block is $Q = 50 \% \text{ of } E = 0.5 \times 1.5 \times 10^6 = 7.5 \times 10^5 \, J$.
The mass of the block $m = 8 \, kg = 8000 \, g$.
The specific heat capacity of aluminium $s = 0.91 \, J/g \cdot ^\circ C$.
Using the formula $Q = ms \Delta T$,we have:
$7.5 \times 10^5 = 8000 \times 0.91 \times \Delta T$.
$7.5 \times 10^5 = 7280 \times \Delta T$.
$\Delta T = \frac{750000}{7280} \approx 103.02 \, ^\circ C$.
Therefore,the rise in temperature is approximately $103 \, ^\circ C$.

(A) Let us calculate the heat required to convert $1\; g$ of ice at $0^{\circ}C$ to water at $100^{\circ}C$:
Heat to melt ice: $Q_1 = m L_f = 1\; g \times 80\; cal/g = 80\; cal$.
Heat to raise water temperature from $0^{\circ}C$ to $100^{\circ}C$: $Q_2 = m c \Delta T = 1\; g \times 1\; cal/g^{\circ}C \times 100^{\circ}C = 100\; cal$.
Total heat required by ice: $Q_{total} = 80 + 100 = 180\; cal$.
Now,calculate the heat released by $1\; g$ of steam at $100^{\circ}C$ to condense into water at $100^{\circ}C$:
Heat released: $Q_{steam} = m L_v = 1\; g \times 540\; cal/g = 540\; cal$.
Since the heat released by steam $(540\; cal)$ is greater than the heat required by ice $(180\; cal)$,the final mixture will be at $100^{\circ}C$ with some steam remaining condensed as water.

(C) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.
Given mass $m = 10\,g = 0.01\,kg$,velocity $v = 300\,m/s$,and specific heat $c = 150\,J/kg\cdot K$.
The total heat produced is equal to the kinetic energy lost: $Q = \frac{1}{2}mv^2$.
Only $50\,\%$ of this heat is absorbed by the bullet,so the heat absorbed is $Q_{abs} = 0.5 \times \frac{1}{2}mv^2 = \frac{1}{4}mv^2$.
The heat absorbed also equals $mc\Delta\theta$,where $\Delta\theta$ is the change in temperature.
Equating the two: $mc\Delta\theta = \frac{1}{4}mv^2$.
$\Delta\theta = \frac{v^2}{4c} = \frac{(300)^2}{4 \times 150} = \frac{90000}{600} = 150^{\circ}C$.

(C) Water exhibits anomalous expansion between $0^{\circ}C$ and $4^{\circ}C$.
When water is heated from $0^{\circ}C$ to $4^{\circ}C$,its volume decreases due to the breaking of hydrogen-bonded clusters,meaning $\Delta V < 0$.
The relationship between molar heat capacities is given by $C_p - C_v = P \left( \frac{\partial V}{\partial T} \right)_p$.
Since the volume $V$ decreases as temperature $T$ increases in this specific range,the derivative $\left( \frac{\partial V}{\partial T} \right)_p$ is negative.
Therefore,$C_p - C_v < 0$,which implies $C_p < C_v$.

(B) According to the principle of conservation of energy,the potential energy lost by the water is converted into heat energy.
$mgh = ms\Delta T$
Here,$m$ is the mass of water,$g$ is the acceleration due to gravity $(9.8\,m/s^2)$,$h$ is the height $(500\,m)$,$s$ is the specific heat capacity of water $(4200\,J/kg\cdot K)$,and $\Delta T$ is the rise in temperature.
$\Delta T = \frac{gh}{s}$
$\Delta T = \frac{9.8 \times 500}{4200}$
$\Delta T = \frac{4900}{4200} = 1.166... \approx 1.16^{\circ}C$

(A) The volume of mercury spilled is given by the difference in the expansion of mercury and the expansion of the glass flask.
The formula for the volume of liquid spilled is $\Delta V = V_0 (\gamma_m - \gamma_g) \Delta T$,where $\gamma_g = 3 \alpha_g$.
Given:
$V_0 = 1 \text{ litre} = 1000 \text{ ml}$
$\gamma_m = 1.82 \times 10^{-4} /^{\circ}C$
$\alpha_g = 10 \times 10^{-6} /^{\circ}C \implies \gamma_g = 3 \times 10 \times 10^{-6} = 0.3 \times 10^{-4} /^{\circ}C$
$\Delta T = 100 - 0 = 100^{\circ}C$
Substituting the values:
$\Delta V = 1000 \times (1.82 \times 10^{-4} - 0.3 \times 10^{-4}) \times 100$
$\Delta V = 1000 \times (1.52 \times 10^{-4}) \times 100$
$\Delta V = 1000 \times 0.0152 = 15.2 \text{ ml}$.

(A) Let $m\, g$ be the mass of steam that condenses into water.
Heat gained by $22\, g$ of water to raise its temperature from $20^{\circ}C$ to $90^{\circ}C$ is:
$Q_{gain} = m_{water} \times c \times \Delta T = 22 \times 1 \times (90 - 20) = 22 \times 70 = 1540\, cal$.
Heat lost by $m\, g$ of steam at $100^{\circ}C$ to become water at $90^{\circ}C$ consists of two parts:
$1$. Heat released during condensation: $Q_1 = m \times L = m \times 540$.
$2$. Heat released by condensed water cooling from $100^{\circ}C$ to $90^{\circ}C$: $Q_2 = m \times c \times \Delta T = m \times 1 \times (100 - 90) = 10m$.
Equating heat lost and heat gained:
$540m + 10m = 1540$
$550m = 1540$
$m = \frac{1540}{550} = 2.8\, g$.
The total mass of water present in the mixture is the initial mass plus the condensed mass:
$M_{total} = 22 + 2.8 = 24.8\, g$.

(D) Water exhibits anomalous expansion. The density of water is maximum at $4\,^{\circ}C$.
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 behavior),causing the volume to increase,which again leads to overflow.
Therefore,water overflows in both cases.

(A) According to the Clausius-Clapeyron equation,the slope of the melting curve is given by $\frac{dP}{dT} = \frac{L}{T(V_2 - V_1)}$,where $L$ is the latent heat of fusion,$T$ is the temperature,$V_2$ is the volume of the liquid phase,and $V_1$ is the volume of the solid phase.
For water,ice contracts upon melting,meaning the volume of water $(V_2)$ is less than the volume of ice $(V_1)$.
Therefore,$(V_2 - V_1) < 0$,which makes the slope $\frac{dP}{dT}$ negative.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(C) The heat supplied at a constant rate is given by $Q = K t$,where $K$ is the heating rate.
For a substance in a single phase,the temperature change is given by $Q = m c \Delta T$,which implies $K t = m c (T - T_0)$. Thus,$T = T_0 + (K / mc) t$. This shows that the temperature increases linearly with time during the heating of a single phase.
When the substance undergoes a phase change (e.g.,vaporization),the heat supplied is used to change the state at a constant temperature,given by $Q = m L$. During this process,the temperature remains constant,resulting in a horizontal line on the $T-t$ graph.
Since oxygen at $1$ atmospheric pressure undergoes a phase change (boiling) between $50\, K$ and $300\, K$,the graph will show a linear increase,followed by a horizontal segment during phase change,and then another linear increase.
Therefore,the correct graph is $C$.

(A) The total heat $Q$ required is the sum of heat for four processes:
$1$. Heating ice from $-12 \; ^{\circ}C$ to $0 \; ^{\circ}C$: $Q_{1} = m \cdot s_{\text{ice}} \cdot \Delta T_{1} = 3 \; kg \times 2100 \; J \; kg^{-1} \; K^{-1} \times (0 - (-12)) \; K = 75600 \; J$.
$2$. Melting ice at $0 \; ^{\circ}C$ to water at $0 \; ^{\circ}C$: $Q_{2} = m \cdot L_{\text{fusion}} = 3 \; kg \times 3.35 \times 10^{5} \; J \; kg^{-1} = 1005000 \; J$.
$3$. Heating water from $0 \; ^{\circ}C$ to $100 \; ^{\circ}C$: $Q_{3} = m \cdot s_{\text{water}} \cdot \Delta T_{2} = 3 \; kg \times 4186 \; J \; kg^{-1} \; K^{-1} \times (100 - 0) \; K = 1255800 \; J$.
$4$. Converting water at $100 \; ^{\circ}C$ to steam at $100 \; ^{\circ}C$: $Q_{4} = m \cdot L_{\text{steam}} = 3 \; kg \times 2.256 \times 10^{6} \; J \; kg^{-1} = 6768000 \; J$.
Total heat $Q = Q_{1} + Q_{2} + Q_{3} + Q_{4} = 75600 + 1005000 + 1255800 + 6768000 = 9104400 \; J \approx 9.1 \times 10^{6} \; J$.

(A) Mass of the copper block,$m = 2.5\; kg = 2500\; g$.
Change in temperature of the copper block,$\Delta \theta = 500\; ^{\circ}C$.
Specific heat of copper,$C = 0.39\; J\; g^{-1}\; ^{\circ}C^{-1}$.
Heat of fusion of water,$L = 335\; J\; g^{-1}$.
The maximum heat the copper block can lose,$Q = m C \Delta \theta$.
$Q = 2500\; g \times 0.39\; J\; g^{-1}\; ^{\circ}C^{-1} \times 500\; ^{\circ}C = 487500\; J$.
Let $m_1$ be the mass of ice that melts. The heat gained by the ice to melt is $Q = m_1 L$.
$m_1 = \frac{Q}{L} = \frac{487500\; J}{335\; J\; g^{-1}} \approx 1455.22\; g$.
Converting to kilograms,$m_1 \approx 1.45\; kg$.

(D) Initial temperature of the child's body,$T_{1} = 101^{\circ} F$.
Final temperature of the child's body,$T_{2} = 98^{\circ} F$.
Change in temperature,$\Delta T = (101 - 98) \times \frac{5}{9} = 3 \times \frac{5}{9} = \frac{5}{3} ^{\circ} C$.
Time taken to reduce the temperature,$t = 20 \; min$.
Mass of the child,$m = 30 \; kg = 30,000 \; g$.
Specific heat of the human body,$c = 1 \; cal \; g^{-1} \; ^{\circ} C^{-1} = 1000 \; cal \; kg^{-1} \; ^{\circ} C^{-1}$.
Latent heat of evaporation of water,$L = 580 \; cal \; g^{-1}$.
The heat lost by the child's body is $\Delta Q = m c \Delta T$.
$\Delta Q = 30,000 \; g \times 1 \; cal \; g^{-1} \; ^{\circ} C^{-1} \times \frac{5}{3} ^{\circ} C = 50,000 \; cal$.
Let $m_{1}$ be the mass of water evaporated. Since $\Delta Q = m_{1} L$,we have $m_{1} = \frac{\Delta Q}{L}$.
$m_{1} = \frac{50,000}{580} \approx 86.21 \; g$.
The average rate of evaporation is $\frac{m_{1}}{t} = \frac{86.21}{20} \approx 4.31 \; g/min$.
Thus,the average rate of extra evaporation is $4.3 \; g/min$.

(N/A) The $P-T$ phase diagram for $CO_{2}$ is shown in the provided figure.
$(a)$ Point $C$ is the triple point of the $CO_{2}$ phase diagram. At this point,the temperature is $-56.6^{\circ}C$ and the pressure is $5.11\;atm$. At these conditions,the solid,liquid,and vapour phases of $CO_{2}$ co-exist in equilibrium.
$(b)$ The fusion and boiling points of $CO_{2}$ decrease with a decrease in pressure.
$(c)$ The critical temperature and critical pressure of $CO_{2}$ are $31.1^{\circ}C$ and $73\;atm$ respectively. The significance is that $CO_{2}$ cannot be liquefied above the critical temperature,regardless of how much pressure is applied.
$(d)$ Based on the $P-T$ phase diagram:
$(i)$ At $-70^{\circ}C$ and $1\;atm$,$CO_{2}$ is in the vapour (gas) phase.
$(ii)$ At $-60^{\circ}C$ and $10\;atm$,$CO_{2}$ is in the solid phase.
$(iii)$ At $15^{\circ}C$ and $56\;atm$,$CO_{2}$ is in the liquid phase.

(D) When an object is heated,several physical changes can occur:
$1$. Its temperature may increase,which is the most common effect.
$2$. It may undergo thermal expansion,where the dimensions of the object increase.
$3$. It may undergo a change of state (e.g.,solid to liquid or liquid to gas) if the heating continues to the melting or boiling point.

(A) The rate of water flow is $m = 3.0 \, L/min = 3000 \, g/min$ (assuming density of water is $1 \, g/mL$).
The temperature rise is $\Delta T = 77 \, ^{\circ}C - 27 \, ^{\circ}C = 50 \, ^{\circ}C$.
The specific heat capacity of water is $c = 4.2 \, J \cdot g^{-1} \cdot ^{\circ}C^{-1}$.
The total heat required per minute is $\Delta Q = m c \Delta T = 3000 \times 4.2 \times 50 = 6.3 \times 10^{5} \, J/min$.
Given the heat of combustion of the fuel is $H = 4.0 \times 10^{4} \, J/g$,the rate of fuel consumption $R$ is given by $R = \frac{\Delta Q}{H}$.
$R = \frac{6.3 \times 10^{5}}{4.0 \times 10^{4}} = 15.75 \, g/min$.

(N/A) When two bodies at different temperatures $T_1$ and $T_2$ are brought into thermal contact,heat flows from the body at the higher temperature to the body at the lower temperature until thermal equilibrium is achieved. The final equilibrium temperature is equal to the mean temperature $(T_1 + T_2) / 2$ only if the heat capacities of both bodies are equal. If the heat capacities differ,the equilibrium temperature will be closer to the temperature of the body with the higher heat capacity.
$(b)$ $A$ coolant in a chemical or nuclear plant must have a high specific heat capacity. This is because a substance with a high specific heat can absorb a large amount of heat energy for a relatively small rise in temperature. This allows the coolant to effectively remove excess heat from the plant components without its own temperature rising to dangerous levels.
$(c)$ When a car is driven,the friction between the tyres and the road,along with the internal compression of air,increases the temperature of the air inside the tyre. According to Gay-Lussac's Law (or Charles' Law for a fixed volume),for a fixed mass of gas in a constant volume,the pressure is directly proportional to the absolute temperature. Thus,as the temperature increases,the air pressure inside the tyre increases.
$(d)$ $A$ harbour town is located near a large body of water,which has a high specific heat capacity. Water heats up and cools down much more slowly than land. Consequently,the water acts as a heat reservoir,moderating the temperature of the air in the harbour town. In contrast,desert sand has a low specific heat capacity,leading to rapid temperature fluctuations,resulting in a more extreme climate.

(A) Before the modern understanding of thermodynamics,heat was believed to be an invisible,weightless fluid called 'caloric'.
This theory,known as the Caloric theory,suggested that heat could flow from hotter bodies to colder bodies and that it was conserved during physical processes.
It was later disproven by experiments conducted by Benjamin Thompson (Count Rumford) and James Prescott Joule,which showed that heat is a form of energy transfer rather than a substance.

(D) The energy released $(Q)$ during cooling is given by the formula $Q = mc\Delta T$,where $m$ is the mass,$c$ is the specific heat capacity of water,and $\Delta T$ is the change in temperature.
Here,$m = 1\,kg$ and the final temperature $T_f = 10\,^{\circ}C$.
The initial temperature $T_i$ is not specified in the question.
Therefore,the energy released $Q = 1 \times 4186 \times (T_i - 10)\,J$.
Since $T_i$ is unknown,the exact amount of energy cannot be calculated.
Thus,the answer depends on the initial temperature of the water.

(N/A) For Solids: $A$ metallic lid on a glass jar that is stuck can be opened by placing it in hot water. The heat causes the metallic lid to expand more than the glass,loosening it for easy removal.
For Liquids: The mercury in a clinical thermometer rises when placed in warm water due to thermal expansion. When the thermometer is removed and cooled,the mercury level falls due to contraction.
For Gases: $A$ partially inflated balloon placed in warm water will expand as the air inside gains kinetic energy and occupies more volume. Conversely,a fully inflated balloon placed in cold water will shrink as the air inside contracts.

(N/A) The boiling point of water is increased inside a pressure cooker by increasing the internal pressure.
In a pressure cooker,as the pressure increases,the boiling point of water rises above its normal value of $100 \ ^\circ\text{C}$.
Consequently,the food is cooked at a higher temperature,which provides more heat energy to the food,resulting in a faster cooking process.

(N/A) Sublimation is the process in which a substance changes directly from the solid state to the vapour state without passing through the intermediate liquid state.
Conversely,the process of changing directly from vapour to solid is also often referred to as sublimation or deposition.
Examples of substances that undergo sublimation include dry ice (solid $CO_{2}$),iodine $(I_{2})$,naphthalene $(C_{10}H_{8})$,and camphor.
During the sublimation process,the solid and vapour phases of the substance coexist in thermal equilibrium at a specific temperature and pressure.

(N/A) Freezing is the phase transition process in which a substance changes from a liquid state to a solid state. This occurs when the thermal energy of the molecules decreases,causing them to lose their mobility and arrange themselves into a fixed,ordered structure.
The freezing point is the specific temperature at which a liquid turns into a solid at a given pressure. At this temperature,the liquid and solid phases of the substance exist in thermal equilibrium. For pure water at standard atmospheric pressure $(1.013 \times 10^5 \ Pa)$,the freezing point is $0 \ ^\circ C$ or $273.15 \ K$.

(B) The main difference lies in the slope of the fusion curve (solid-liquid equilibrium line).
For water,the fusion curve has a negative slope because the density of ice is less than that of liquid water,meaning an increase in pressure lowers the melting point.
For $CO_2$,the fusion curve has a positive slope because the solid phase is denser than the liquid phase,meaning an increase in pressure raises the melting point.

(N/A) $1.$ The freezing point of water is $32^{\circ}F$ and the boiling point is $212^{\circ}F$ on a Fahrenheit thermometer.
$2.$ The zeroth law of thermodynamics defines temperature,and the first law of thermodynamics defines internal energy.
$3.$ At the critical temperature $(647.096 \ K)$,water and water vapour have equal density.
$4.$ Water has the maximum value of specific heat capacity among common substances.

(N/A) The Stefan-Boltzmann law is given by $E = \sigma T^4$. Thus,the unit of $\sigma$ is $W \cdot m^{-2} \cdot K^{-4}$.
$(b)$ The temperature gradient is defined as $\frac{dT}{dx} = 5\,^{\circ}C/cm$. The temperature difference $\Delta T = 100\,^{\circ}C - 60\,^{\circ}C = 40\,^{\circ}C$. The distance $x$ is given by $x = \frac{\Delta T}{dT/dx} = \frac{40}{5} = 8\,cm$.
$(c)$ The coefficient of volume expansion of water is zero at $4\,^{\circ}C$ because water has maximum density at this temperature.

(A) In a phase diagram of a substance:
$1$. The Vaporization curve represents the boundary where liquid and gaseous phases coexist in equilibrium.
$2$. The Sublimation curve represents the boundary where solid and gaseous phases coexist in equilibrium.
$3$. The Fusion curve represents the boundary where solid and liquid phases coexist in equilibrium.
Therefore,$(a)$ matches with $(iii)$ and $(b)$ matches with $(i)$.
The correct option is $(a-iii), (b-i)$.

(N/A) The $p-T$ phase diagram of water shows that the fusion curve (the boundary between solid and liquid phases) has a negative slope. This means that for water,an increase in pressure at a constant temperature near $0^{\circ}C$ lowers the melting point of ice.
When crushed ice or snow is squeezed in the palm,the pressure applied at the contact points between the ice particles increases. Due to the negative slope of the fusion curve,this increased pressure causes the ice to melt locally,even if the temperature is slightly below $0^{\circ}C$.
As the pressure is released (when the hand is opened or the pressure is redistributed),the local pressure drops back to atmospheric pressure. Consequently,the water that was formed by melting refreezes,acting as a binder that fuses the individual ice particles together into a solid,stable ball.

(B) The initial kinetic energy of the bullet is $K = \frac{1}{2} m v^2$.
Given $m = 5\, g$ and $v = 210\, m/s = 21000\, cm/s$.
$K = \frac{1}{2} \times 5 \times (21000)^2 = 1.1025 \times 10^9\, ergs$.
One half of this energy is converted into heat in the bullet:
$Q = \frac{1}{2} K = \frac{1}{2} \times 1.1025 \times 10^9 = 5.5125 \times 10^8\, ergs$.
We know $Q = m s \Delta T$,where $s = 0.030\, cal/(g \cdot ^{\circ}C) = 0.030 \times 4.2 \times 10^7\, ergs/(g \cdot ^{\circ}C) = 1.26 \times 10^6\, ergs/(g \cdot ^{\circ}C)$.
Substituting the values:
$5.5125 \times 10^8 = 5 \times (1.26 \times 10^6) \times \Delta T$.
$\Delta T = \frac{5.5125 \times 10^8}{6.3 \times 10^6} = \frac{551.25}{6.3} = 87.5^{\circ}C$.

(A) The potential energy lost by the water as it falls is converted into heat energy,which increases the temperature of the water.
Change in potential energy $(P.E.)$ = Heat energy $(Q)$
$mgh = mS \Delta T$
$\Delta T = \frac{gh}{S}$
Given:
$g = 10 \ m/s^2$
$h = 63 \ m$
$S = 1 \ cal \ g^{-1} \ ^{\circ}C^{-1} = 1 \times 4.2 \ J \ g^{-1} \ ^{\circ}C^{-1} = 4200 \ J \ kg^{-1} \ ^{\circ}C^{-1}$
Substituting the values:
$\Delta T = \frac{10 \times 63}{4200}$
$\Delta T = \frac{630}{4200} = \frac{63}{420} = 0.15 \approx 0.147 \ ^{\circ}C$
Thus,the difference in temperature is $0.147 \ ^{\circ}C$.

(C) The kinetic energy of the hammer is given by $K.E. = \frac{1}{2} mv^2 = \frac{1}{2} \times 1.5 \times (60)^2 = 0.75 \times 3600 = 2700 \, J$.
According to the problem,one-fourth of this energy is used to heat the iron nail.
Heat energy absorbed by the nail $Q = \frac{1}{4} \times 2700 = 675 \, J$.
The formula for heat absorbed is $Q = mc\Delta T$,where $m = 100 \, g$,$c = 0.42 \, Jg^{-1} {}^{\circ}C^{-1}$,and $\Delta T$ is the rise in temperature.
Substituting the values: $675 = 100 \times 0.42 \times \Delta T$.
$675 = 42 \times \Delta T$.
$\Delta T = \frac{675}{42} \approx 16.07 \, ^{\circ}C$.

(C) The heat rejected consists of two parts: the heat released during condensation and the heat released during cooling of water.
$1$. Heat released during condensation $(Q_1)$: $Q_1 = m \times L_v = 50 \, g \times 540 \, cal/g = 27000 \, cal$.
$2$. Heat released during cooling of water from $100^{\circ} C$ to $20^{\circ} C$ $(Q_2)$: $Q_2 = m \times s \times \Delta T = 50 \, g \times 1 \, cal/g^{\circ} C \times (100^{\circ} C - 20^{\circ} C) = 50 \times 80 = 4000 \, cal$.
Total heat rejected $(Q_{total})$ = $Q_1 + Q_2 = 27000 + 4000 = 31000 \, cal$.
$Q_{total} = 31 \times 10^{3} \, cal$.

(B) The mass flow rate of water is $m = 2.0 \; kg/min = 2000 \; g/min$.
The temperature change is $\Delta T = 70^{\circ} C - 30^{\circ} C = 40^{\circ} C$.
The specific heat capacity of water is $S = 4.2 \; J \cdot g^{-1} \cdot {}^{\circ} C^{-1}$.
The heat required per minute is $Q = m \cdot S \cdot \Delta T$.
$Q = 2000 \times 4.2 \times 40 = 336000 \; J/min$.
Let the rate of combustion of fuel be $R$ (in $g/min$) and the heat of combustion be $L = 8 \times 10^{3} \; J/g$.
The heat supplied by the fuel is $Q = R \times L$.
$336000 = R \times 8 \times 10^{3}$.
$R = \frac{336000}{8000} = 42 \; g/min$.

(A) The correct answer is $(A)$.
On the surface of the moon,there is no atmosphere,which means the external atmospheric pressure is $0$.
$A$ liquid boils when its saturated vapour pressure becomes equal to the external atmospheric pressure.
Since the vapour pressure of water at $30^{\circ} C$ is greater than $0$,the condition for boiling is satisfied immediately upon opening the bottle.
Therefore,the water will start to boil as soon as the bottle is opened on the moon.

(D) The change in volume $\Delta V$ of a liquid is given by $\Delta V = V_0 \gamma \Delta \theta$,where $V_0$ is the initial volume,$\gamma$ is the coefficient of volume expansion,and $\Delta \theta$ is the change in temperature.
Since the increase in length $\Delta l$ is the same for both thermometers,the change in volume is $\Delta V = A \Delta l = \frac{\pi d^2}{4} \Delta l$.
Given that the initial volumes $(V_0)_{Hg} = (V_0)_{Br}$ and the change in length $\Delta l$ is the same for both,we have:
$\Delta V_{Hg} = \frac{\pi d_1^2}{4} \Delta l = V_0 \gamma_{Hg} \Delta \theta$
$\Delta V_{Br} = \frac{\pi d_2^2}{4} \Delta l = V_0 \gamma_{Br} \Delta \theta$
Dividing the two equations:
$\frac{d_1^2}{d_2^2} = \frac{\gamma_{Hg}}{\gamma_{Br}}$
$\frac{d_1}{d_2} = \sqrt{\frac{\gamma_{Hg}}{\gamma_{Br}}}$
Substituting the given values:
$\frac{d_1}{d_2} = \sqrt{\frac{18 \times 10^{-5}}{108 \times 10^{-5}}} = \sqrt{\frac{1}{6}} \approx \sqrt{0.166} \approx 0.408$
Thus,the ratio $d_1: d_2$ is closest to $0.4$.

(C) Mass of water evaporated in two hours,$m = 2 \,h \times 20 \,g/h = 40 \,g = 40 \times 10^{-3} \,kg$.
Heat absorbed by water during evaporation is $Q = m L$,where $L$ is the latent heat of vaporization.
Assuming this heat is taken entirely from the remaining water in the earthen pot,the heat lost by the water is $Q = M s \Delta T$,where $M = 4 \,kg$ is the mass of water and $s$ is the specific heat of water.
Equating the heat absorbed and heat lost: $m L = M s \Delta T$.
Therefore,$\Delta T = \frac{m}{M} \times \frac{L}{s}$.
Given $\frac{L}{s} = 540^{\circ} C$,we have $\Delta T = \frac{40 \times 10^{-3}}{4} \times 540$.
$\Delta T = 0.01 \times 540 = 5.4^{\circ} C$.

(B) Let $P$ be the rate of heat removal (power). Since heat is extracted at the same rate,$P$ is constant for both.
The heat lost by a substance is given by $dQ = m \cdot c \cdot dT$,where $m$ is the mass and $c$ is the specific heat.
The rate of heat removal is $P = \frac{|dQ|}{dt} = m \cdot c \cdot \left| \frac{dT}{dt} \right|$.
Thus,the magnitude of the slope of the $T-t$ graph is $\left| \frac{dT}{dt} \right| = \frac{P}{m \cdot c}$.
Since $P$ and $m$ are constant,the slope is inversely proportional to the specific heat: $\text{Slope} \propto \frac{1}{c}$.
$1$. For the liquid state (the initial cooling phase): Looking at the graph,the slope of curve $1$ is smaller than the slope of curve $2$. Since $\text{Slope} \propto \frac{1}{C_L}$,a smaller slope implies a larger specific heat. Therefore,$C_{L1} > C_{L2}$.
$2$. For the solid state (the final cooling phase): Looking at the graph,the slope of curve $1$ is larger than the slope of curve $2$. Since $\text{Slope} \propto \frac{1}{C_S}$,a larger slope implies a smaller specific heat. Therefore,$C_{S1} < C_{S2}$.
Thus,the correct option is $C_{L1} > C_{L2}$ and $C_{S1} < C_{S2}$.

(C) The kinetic energy of the bullet is converted into heat,which increases the temperature of both the bullet and the wax.
By the principle of energy conservation:
$\frac{1}{2} m_b v_b^2 = (m_w c_w + m_b c_b) \Delta T$
Given:
$m_b = 20 \,g = 0.02 \,kg$,$v_b = 2000 \,m/s$,$c_b = 5000 \,J/(kg \cdot ^{\circ}C)$
$m_w = 1.0 \,kg$,$c_w = 3000 \,J/(kg \cdot ^{\circ}C)$,$T_i = 25^{\circ}C$
Substituting the values:
$\frac{1}{2} \times 0.02 \times (2000)^2 = (1.0 \times 3000 + 0.02 \times 5000) \Delta T$
$0.01 \times 4,000,000 = (3000 + 100) \Delta T$
$40,000 = 3100 \Delta T$
$\Delta T = \frac{400}{31} \approx 12.9^{\circ}C$
Final temperature $T_f = T_i + \Delta T = 25 + 12.9 = 37.9^{\circ}C$.

(A) Let $C$ be the heat capacity of the container $(C = m_c s_c)$.
Case $1$: Water in the container.
Energy supplied = $P \times t_1 = 10 \times (15 \times 60) = 9000 \,J$.
Heat absorbed = $(m_w s_w + C) \Delta T_1 = (0.5 \times 4200 + C) \times 3 = 6300 + 3C$.
Equating energy: $6300 + 3C = 9000 \Rightarrow 3C = 2700 \Rightarrow C = 900 \,J K^{-1}$.
Case $2$: Oil in the container.
Energy supplied = $P \times t_2 = 10 \times (20 \times 60) = 12000 \,J$.
Heat absorbed = $(m_o s_o + C) \Delta T_2 = (2 \times s_o + 900) \times 2 = 4s_o + 1800$.
Equating energy: $4s_o + 1800 = 12000 \Rightarrow 4s_o = 10200 \Rightarrow s_o = 2550 \,J K^{-1} kg^{-1}$.
Expressing in terms of $10^3 \,J K^{-1} kg^{-1}$,we get $s_o = 2.55 \times 10^3 \,J K^{-1} kg^{-1}$. Given the options,the closest value is $2.5$.

(C) The process occurs in three stages:
$1$. Heating ice from $-8^{\circ} C$ to $0^{\circ} C$: $Q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \,kg \times 2.1 \,kJ/kg \cdot K \times 8 \,K = 16.8 \,kJ$.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: $Q_2 = m \cdot L_f = 1 \,kg \times 333 \,kJ/kg = 333 \,kJ$.
$3$. Heating water from $0^{\circ} C$ to $20^{\circ} C$: $Q_3 = m \cdot c_{water} \cdot \Delta T = 1 \,kg \times 4.2 \,kJ/kg \cdot K \times 20 \,K = 84 \,kJ$.
Total heat required $Q = Q_1 + Q_2 + Q_3 = 16.8 + 333 + 84 = 433.8 \,kJ$.
Rounding to the nearest integer,we get $434 \,kJ$.

(B) The volume of the flask $V_0 = 200 \, cm^3$ and the change in temperature $\Delta T = 100^{\circ} C - 20^{\circ} C = 80^{\circ} C$.
The expansion of the glass flask is given by $\Delta V_{\text{glass}} = V_0 \gamma_{\text{glass}} \Delta T$.
$\Delta V_{\text{glass}} = 200 \times (1.2 \times 10^{-5}) \times 80 = 0.192 \, cm^3$.
The expansion of the mercury is given by $\Delta V_{\text{mercury}} = V_0 \gamma_{\text{mercury}} \Delta T$.
$\Delta V_{\text{mercury}} = 200 \times (1.8 \times 10^{-4}) \times 80 = 2.88 \, cm^3$.
The amount of mercury that overflows is the difference between the expansion of mercury and the expansion of the glass flask:
$\Delta V_{\text{overflow}} = \Delta V_{\text{mercury}} - \Delta V_{\text{glass}} = 2.88 - 0.192 = 2.688 \, cm^3 \approx 2.69 \, cm^3$.

(A) The total heat required is the sum of heat for four processes:
$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: $Q_1 = m c_{ice} \Delta T = 1 \times 2.1 \times 10 = 21 \, J$.
$2$. Melting ice at $0^{\circ} C$: $Q_2 = m L_f = 1 \times 334 = 334 \, J$.
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m c_{water} \Delta T = 1 \times 4.18 \times 100 = 418 \, J$.
$4$. Converting water to steam at $100^{\circ} C$: $Q_4 = m L_v = 1 \times 2260 = 2260 \, J$.
Total heat $Q = Q_1 + Q_2 + Q_3 + Q_4 = 21 + 334 + 418 + 2260 = 3033 \, J$.
Converting to $kJ$: $Q = 3.033 \, kJ \approx 3.04 \, kJ$.

(C) The correct option is $C$.
Water exhibits anomalous expansion,meaning its density is maximum at $4^{\circ} C$.
In a deep lake,the water at the bottom remains at the temperature where its density is highest to maintain stability.
Therefore,regardless of the surface temperature (as long as it is below $4^{\circ} C$),the temperature at the bottom of the lake will be $4^{\circ} C$.

(C) Water exhibits anomalous expansion. The density of water is maximum at $4^{\circ} C$.
In winter,as the surface of the lake cools down,the water at the surface becomes denser and sinks to the bottom.
This process continues until the entire body of water reaches $4^{\circ} C$.
If the surface temperature drops further below $4^{\circ} C$,the water at the surface becomes less dense and stays at the top.
Therefore,the water at the bottom remains at $4^{\circ} C$ while the surface may freeze or reach lower temperatures.
Thus,the temperature at the bottom of the lake is $4^{\circ} C$.

(A) The process of heating ice from $-12^{\circ} C$ to steam at $100^{\circ} C$ involves several stages:
$1$. Heating ice from $-12^{\circ} C$ to $0^{\circ} C$: The temperature increases linearly with heat added $(Q = mc\Delta T)$.
$2$. Melting ice at $0^{\circ} C$: The temperature remains constant at $0^{\circ} C$ while the phase changes from solid to liquid $(Q = mL_f)$.
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: The temperature increases linearly with heat added $(Q = mc\Delta T)$.
$4$. Boiling water at $100^{\circ} C$: The temperature remains constant at $100^{\circ} C$ while the phase changes from liquid to gas $(Q = mL_v)$.
The correct curve must show two sloped regions (temperature rise) and two horizontal regions (phase change at constant temperature). This corresponds to the provided solution image.

(D) The rate of heat supply is given by $\frac{dQ}{dt} = \frac{dm}{dt} \times L$,where $L$ is the latent heat of fusion of ice $(80 \,cal/g)$.
Given $\frac{dm}{dt} = 0.1 \,g/s$,we have $\frac{dQ}{dt} = 0.1 \times 80 = 8 \,cal/s$.
After the ice melts,the total mass of the water formed is $m = \frac{dm}{dt} \times t = 0.1 \,g/s \times 100 \,s = 10 \,g$.
The heat supplied to the water is $Q = ms\Delta T$,where $s$ is the specific heat capacity of water $(1 \,cal/g^{\circ}C)$.
Differentiating with respect to time,we get $\frac{dQ}{dt} = ms \frac{dT}{dt}$.
Substituting the values: $8 = 10 \times 1 \times \frac{dT}{dt}$.
Therefore,$\frac{dT}{dt} = \frac{8}{10} = 0.8 \,^{\circ}C/s$.

(C) According to Newton's law of cooling,the rate of loss of heat is proportional to the temperature difference between the body and the surroundings. Since the temperature range ($60^{\circ}C$ to $55^{\circ}C$) is the same for both,the average rate of heat loss is approximately constant.
Let $m_w = 250\,g$ (mass of water),$m_a = 200\,g$ (mass of alcohol),$W = 10\,g$ (water equivalent of calorimeter),and $s$ be the specific heat of alcohol. The specific heat of water $s_w = 1\,\text{cal/g}^{\circ}C$.
Heat lost by water and calorimeter: $Q_w = (m_w + W) s_w \Delta T = (250 + 10) \times 1 \times 5 = 1300\,\text{cal}$.
Rate of heat loss for water: $R_w = \frac{Q_w}{t_w} = \frac{1300}{130} = 10\,\text{cal/s}$.
Heat lost by alcohol and calorimeter: $Q_a = (m_a s + W) \Delta T = (200s + 10) \times 5 = 1000s + 50\,\text{cal}$.
Rate of heat loss for alcohol: $R_a = \frac{Q_a}{t_a} = \frac{1000s + 50}{67}$.
Since the rate of heat loss is the same in both cases $(R_w = R_a)$:
$10 = \frac{1000s + 50}{67}$
$670 = 1000s + 50$
$1000s = 620$
$s = 0.62\,\text{cal/g}^{\circ}C$.

(A) Given: Heat $\Delta Q = 184 \times 10^3\,J$,mass $m = 0.6\,kg$,initial temperature $T_i = -12^{\circ}C$,specific heat $c_{ice} = 2222.3\,J\,kg^{-1\circ}C^{-1}$,latent heat $L = 336 \times 10^3\,J\,kg^{-1}$.
Step $1$: Heat required to raise the temperature of ice from $-12^{\circ}C$ to $0^{\circ}C$:
$Q_1 = m \cdot c_{ice} \cdot \Delta T = 0.6 \times 2222.3 \times 12 = 16000.56\,J$.
Step $2$: Remaining heat available for melting:
$Q_{rem} = \Delta Q - Q_1 = 184000 - 16000.56 = 167999.44\,J$.
Step $3$: Heat required to melt the entire mass of ice at $0^{\circ}C$:
$Q_{melt} = m \cdot L = 0.6 \times 336000 = 201600\,J$.
Since $Q_{rem} < Q_{melt}$,the ice will not melt completely,and the final temperature will be $0^{\circ}C$. Thus,statement $(A)$ is correct.
Step $4$: Calculate the mass of ice melted $(m_w)$:
$m_w = \frac{Q_{rem}}{L} = \frac{167999.44}{336000} \approx 0.5\,kg$.
Step $5$: Calculate the remaining mass of ice $(m_i)$:
$m_i = m - m_w = 0.6 - 0.5 = 0.1\,kg$.
Step $6$: Ratio of ice to water:
Ratio $= \frac{m_i}{m_w} = \frac{0.1}{0.5} = 1:5$. Thus,statement $(D)$ is correct.
Therefore,$(A)$ and $(D)$ are correct.