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

(B) Given: Resistance at ice point $R_0 = 8 \Omega$,Resistance at steam point $R_{100} = 10 \Omega$.
The formula for resistance at temperature $T$ is $R_T = R_0(1 + \alpha T)$.
For the steam point $(100^{\circ} C)$:
$10 = 8(1 + \alpha \times 100)$
$1.25 = 1 + 100\alpha$
$100\alpha = 0.25 = 1/4$
$\alpha = 1/400 \text{ per } ^{\circ} C$.
Now,for the temperature $T = 400^{\circ} C$:
$R_{400} = R_0(1 + \alpha \times 400)$
$R_{400} = 8(1 + (1/400) \times 400)$
$R_{400} = 8(1 + 1) = 8 \times 2 = 16 \Omega$.
Therefore,the correct option is $B$.

$(A)$ Bimetallic strip works on the principle of thermal expansion of solids, where two different metals expand by different amounts when heated, causing the strip to bend. Thus, $A \rightarrow (s)$.
$(B)$ $A$ steam engine converts thermal energy (from steam) into mechanical work. Thus, $B \rightarrow (q)$.
$(C)$ An incandescent lamp works by heating a filament to a high temperature, causing it to emit light via radiation. Thus, $C \rightarrow (p)$.
$(D)$ An electric fuse is a safety device that melts when the current exceeds a specific limit, breaking the circuit. Thus, $D \rightarrow (r)$.
Therefore, the correct matching is $A \rightarrow (s), B \rightarrow (q), C \rightarrow (p), D \rightarrow (r)$.

(B) Total heat generated by the device in $t = 3 \text{ hours} = 3 \times 3600 \text{ s} = 10800 \text{ s}$ is:
$Q_{\text{gen}} = P_{\text{device}} \times t = 3000 \text{ W} \times 10800 \text{ s} = 3.24 \times 10^7 \text{ J}$.
The heat absorbed by the $120 \text{ litres}$ $(120 \text{ kg})$ of water as its temperature rises from $10^{\circ}\text{C}$ to $30^{\circ}\text{C}$ is:
$Q_{\text{water}} = m \cdot c \cdot \Delta T = 120 \text{ kg} \times 4200 \text{ J kg}^{-1} \text{K}^{-1} \times (30 - 10) \text{ K} = 120 \times 4200 \times 20 = 1.008 \times 10^7 \text{ J}$.
The heat that must be removed by the cooler is:
$Q_{\text{cooler}} = Q_{\text{gen}} - Q_{\text{water}} = 3.24 \times 10^7 \text{ J} - 1.008 \times 10^7 \text{ J} = 2.232 \times 10^7 \text{ J}$.
The minimum power $P$ of the cooler is:
$P = \frac{Q_{\text{cooler}}}{t} = \frac{2.232 \times 10^7 \text{ J}}{10800 \text{ s}} \approx 2066.67 \text{ W} \approx 2067 \text{ W}$.

(C) Let the mass of the calorimeter be $m$,the specific heat of the calorimeter be $x$,the specific heat of the liquid be $s$,and the latent heat of the liquid be $L$.
When $5 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $110^{\circ} C$,the calorimeter cools down to $80^{\circ} C$ and the liquid evaporates:
$m \cdot x \cdot (110 - 80) = 5 \cdot s \cdot (80 - 30) + 5 \cdot L$
$30 \cdot mx = 250s + 5L$ ... $(i)$
Now,$80 \ gm$ of liquid at $30^{\circ} C$ is added to the calorimeter at $80^{\circ} C$ and the final equilibrium temperature becomes $50^{\circ} C$:
$m \cdot x \cdot (80 - 50) = 80 \cdot s \cdot (50 - 30)$
$30 \cdot mx = 1600s$ ... $(ii)$
Comparing equations $(i)$ and $(ii)$:
$250s + 5L = 1600s$
$5L = 1350s$
$\frac{L}{s} = \frac{1350}{5} = 270$

(B) The total heat required is the sum of heat absorbed in five stages:
$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: $\Delta Q_1 = m \cdot S_{\text{ice}} \cdot \Delta T = 10^{-3} \times 2100 \times 10 = 21 \ J$
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: $\Delta Q_2 = m \cdot L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \ J$
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: $\Delta Q_3 = m \cdot S_{\text{water}} \cdot \Delta T = 10^{-3} \times 4180 \times 100 = 418 \ J$
$4$. Vaporizing water at $100^{\circ} C$ to steam at $100^{\circ} C$: $\Delta Q_4 = m \cdot L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \ J$
$5$. Heating steam from $100^{\circ} C$ to $110^{\circ} C$: $\Delta Q_5 = m \cdot S_{\text{steam}} \cdot \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \ J$
Total heat $\Delta Q_{\text{total}} = \Delta Q_1 + \Delta Q_2 + \Delta Q_3 + \Delta Q_4 + \Delta Q_5 = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \ J$.
Rounding to the nearest integer,we get $3043 \ J$.

(C) The total heat required for the process consists of two parts: the heat required to raise the temperature of the bullet to its melting point and the heat required for the phase change (melting).
Let $m$ be the mass of the bullet in $kg$.
Heat required to raise the temperature from $300 \ K$ to $600 \ K$ is $Q_1 = ms \Delta T = m \times 125 \times (600 - 300) = m \times 125 \times 300 = 37500m \ J$.
Heat required for melting is $Q_2 = mL = m \times 2.5 \times 10^4 = 25000m \ J$.
Total heat $Q = Q_1 + Q_2 = 37500m + 25000m = 62500m \ J$.
Given $Q = 625 \ J$,we have $62500m = 625$.
$m = \frac{625}{62500} = 0.01 \ kg$.
Converting to grams,$m = 0.01 \times 1000 = 10 \ g$.

(D) $1$. Heat capacity $(C^{\prime})$ is defined as the amount of heat required to raise the temperature of a body by $1 K$. Its unit is $J K^{-1}$. Thus,$(A)-(II)$.
$2$. Specific heat capacity $(S)$ is the amount of heat required to raise the temperature of $1 kg$ of a substance by $1 K$. Its unit is $J kg^{-1} K^{-1}$. Thus,$(B)-(III)$.
$3$. Latent heat $(L)$ is the heat required to change the state of $1 kg$ of a substance at constant temperature. Its unit is $J kg^{-1}$. Thus,$(C)-(I)$.
$4$. Thermal conductivity $(K)$ is given by the formula $\Delta Q = \frac{KA \Delta T}{L} t$. Rearranging for $K$,we get $K = \frac{\Delta Q \cdot L}{A \cdot \Delta T \cdot t}$. The unit is $J m^{-1} K^{-1} s^{-1}$. Thus,$(D)-(IV)$.
Therefore,the correct matching is $(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$.

(A) The initial area of the sheet is $A_0 = l \times d = 9 \ cm \times 4 \ cm = 36 \ cm^2 = 36 \times 10^{-4} \ m^2$.
Using the heat formula $\Delta Q = m C_v \Delta T$,we find the change in temperature $\Delta T$:
$8.1 \times 10^2 = 0.1 \times 900 \times \Delta T$
$810 = 90 \times \Delta T$
$\Delta T = \frac{810}{90} = 9 \ K$.
The change in area $\Delta A$ due to thermal expansion is given by $\Delta A = A_0 \beta \Delta T$,where $\beta = 2\alpha$ is the coefficient of areal expansion.
$\Delta A = A_0 (2\alpha) \Delta T$
$\Delta A = (36 \times 10^{-4} \ m^2) \times (2 \times 3.1 \times 10^{-5} \ K^{-1}) \times (9 \ K)$
$\Delta A = 36 \times 10^{-4} \times 6.2 \times 10^{-5} \times 9$
$\Delta A = 2008.8 \times 10^{-9} \ m^2 \approx 2.0 \times 10^{-6} \ m^2$.

(D) The potential energy of the water at height $h$ is converted into heat energy when it hits the pool.
$mgh = ms \Delta T$
Here,$m$ is the mass of water,$g = 10 \ m/s^2$ is the acceleration due to gravity,$h = 200 \ m$ is the height,and $s = 4200 \ J/(kg \ K)$ is the specific heat capacity of water.
Canceling $m$ from both sides,we get:
$gh = s \Delta T$
$\Delta T = \frac{gh}{s}$
Substituting the values:
$\Delta T = \frac{10 \times 200}{4200} = \frac{2000}{4200} = \frac{20}{42} = \frac{10}{21} \ K$
$\Delta T \approx 0.476 \ K \approx 0.48 \ K$

(B) The kinetic energy of the bullet is $K.E. = \frac{1}{2} mv^2$.
Given: $m = 20 \ g = 0.02 \ kg$,$v = 300 \ m/s$,$S = 150 \ J/kg \cdot K$.
$K.E. = \frac{1}{2} \times 0.02 \times (300)^2 = 0.01 \times 90000 = 900 \ J$.
Since $50 \%$ of the heat is absorbed by the bullet,the heat absorbed $Q = 0.5 \times 900 = 450 \ J$.
Using the formula $Q = mS \Delta T$,we have $450 = 0.02 \times 150 \times \Delta T$.
$450 = 3 \times \Delta T$.
$\Delta T = \frac{450}{3} = 150^{\circ} C$.

(C) The potential energy of the object at height $h$ is converted into heat energy upon impact.
Given: Mass $m = 3.5 \ kg$,height $h = 2000 \ m$,$g = 10 \ m/s^2$,Latent heat $L = 3.5 \times 10^5 \ J/kg$.
The potential energy $PE = mgh = 3.5 \times 10 \times 2000 = 70,000 \ J$.
The heat required to melt a mass $m'$ of ice is $Q = m'L$.
Since the object comes to rest,all potential energy is converted into heat: $mgh = m'L$.
$70,000 = m' \times 3.5 \times 10^5$.
$m' = \frac{70,000}{3.5 \times 10^5} = \frac{7 \times 10^4}{3.5 \times 10^5} = 2 \times 10^{-1} \ kg = 0.2 \ kg$.
Converting to grams: $0.2 \ kg = 200 \ g$.

(B) The kinetic energy of the bullet is $K.E. = \frac{1}{2} mv^2$.
Given that $50\%$ of this energy is converted into heat energy $(Q)$.
Therefore,the heat energy produced is $Q = \frac{1}{2} \times (\frac{1}{2} mv^2) = \frac{1}{4} mv^2$.
The heat required to raise the temperature of the bullet by $\Delta T$ is given by $Q = ms \Delta T$.
Since $J$ is the mechanical equivalent of heat,we use the relation $W = JQ$,where $W$ is the work done (or energy converted).
Equating the energy converted to heat: $\frac{1}{4} mv^2 = J(ms \Delta T)$.
Solving for $\Delta T$: $\Delta T = \frac{mv^2}{4 Jms} = \frac{v^2}{4 Js}$.

(D) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the derivative,the change in time period is $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.
Since $\frac{dL}{L} = \alpha \Delta \theta$,we have $\frac{dT}{T} = \frac{1}{2} \alpha \Delta \theta$.
Here,$T = 0.5 \ s$,$\alpha = 9 \times 10^{-7} /^{\circ}C$,and $\Delta \theta = 30^{\circ}C - 20^{\circ}C = 10^{\circ}C$.
Substituting the values: $dT = T \times \frac{1}{2} \times \alpha \times \Delta \theta$.
$dT = 0.5 \times \frac{1}{2} \times (9 \times 10^{-7}) \times 10$.
$dT = 0.25 \times 9 \times 10^{-6} = 2.25 \times 10^{-6} \ s$.

(A) The initial kinetic energy of the bullet is $K.E. = \frac{1}{2} MV^2$.
Given that $75 \%$ of this energy is converted into heat,the heat energy produced is $Q = 0.75 \times K.E. = \frac{3}{4} \times \frac{1}{2} MV^2 = \frac{3}{8} MV^2$.
The heat energy required to raise the temperature of the bullet by $\Delta T$ is given by $H = M s \Delta T$.
Using the mechanical equivalent of heat $J$,the relationship between heat energy in Joules $(W)$ and calories $(Q)$ is $W = JQ$.
Here,the work done (energy converted) is $W = \frac{3}{8} MV^2$ and the heat absorbed is $Q = M s \Delta T$.
Thus,$\frac{3}{8} MV^2 = J (M s \Delta T)$.
Solving for the increase in temperature $\Delta T$:
$\Delta T = \frac{3 MV^2}{8 J M s} = \frac{3 V^2}{8 Js}$.

(B) Heat required to melt $ 1 \text{ g} $ of ice at $ 0^{\circ} C $ to water at $ 0^{\circ} C $ is $ Q_1 = m L_f = 1 \times 80 = 80 \text{ cal} $.
Heat required to raise the temperature of $ 1 \text{ g} $ of water from $ 0^{\circ} C $ to $ 100^{\circ} C $ is $ Q_2 = m c \Delta T = 1 \times 1 \times 100 = 100 \text{ cal} $.
Total heat required to convert $ 1 \text{ g} $ of ice at $ 0^{\circ} C $ to $ 1 \text{ g} $ of water at $ 100^{\circ} C $ is $ Q_{total} = 80 + 100 = 180 \text{ cal} $.
Heat released by $ 1 \text{ g} $ of steam at $ 100^{\circ} C $ to condense into water at $ 100^{\circ} C $ is $ Q_{steam} = m L_v = 1 \times 540 = 540 \text{ cal} $.
Since $ Q_{steam} > Q_{total} $,the steam has more than enough heat to melt the ice and raise the water to $ 100^{\circ} C $.
Therefore,the mixture will reach thermal equilibrium at $ 100^{\circ} C $ with some steam remaining.

(C) The total heat energy $Q$ required to convert the solid cube at $\theta_0$ to vapour at $\theta_2$ is the sum of heat required for each phase change and temperature change step:
$1$. Heat to raise temperature of solid from $\theta_0$ to $\theta_1$: $Q_1 = m s_1(\theta_1 - \theta_0)$
$2$. Heat for fusion at $\theta_1$: $Q_2 = m L_f$
$3$. Heat to raise temperature of liquid from $\theta_1$ to $\theta_2$: $Q_3 = m s_2(\theta_2 - \theta_1)$
$4$. Heat for vaporisation at $\theta_2$: $Q_4 = m L_v$
Total heat $Q = Q_1 + Q_2 + Q_3 + Q_4 = m s_1(\theta_1 - \theta_0) + m L_f + m s_2(\theta_2 - \theta_1) + m L_v$.

(D) The total heat supplied $Q$ is given by the formula: $Q = (m_1 s_1 + m_2 s_2 + m_3 s_3) \Delta T$.
Here,$m_1 = 2.4 \ kg$,$s_1 = 0.216 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Aluminium).
$m_2 = 1.6 \ kg$,$s_2 = 0.0917 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Brass).
$m_3 = 0.8 \ kg$,$s_3 = 0.0931 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ (Copper).
$Q = 44.4 \ cal$.
Substituting the values: $44.4 = (2.4 \times 0.216 + 1.6 \times 0.0917 + 0.8 \times 0.0931) \Delta T$.
$44.4 = (0.5184 + 0.14672 + 0.07448) \Delta T$.
$44.4 = (0.7396) \Delta T$.
$\Delta T = 44.4 / 0.7396 \approx 60^{\circ} C$.
Since $\Delta T = T_{final} - T_{initial}$,we have $60 = T_{final} - 20$.
Therefore,$T_{final} = 80^{\circ} C$.

(B) Given: Mass $m = 2 \,kg$, inclination $\theta = \sin^{-1}(3/5)$, so $\sin \theta = 3/5$ and $\cos \theta = 4/5$. Distance $s = 80 \,cm = 0.8 \,m$. Specific heat $c = 420 \,J kg^{-1} K^{-1}$. Acceleration due to gravity $g = 10 \,ms^{-2}$.
Since the block slides at a constant speed, the work done by friction equals the loss in potential energy. The work done by friction $W_f = f_k \cdot s$, where $f_k = \mu_k N = \mu_k mg \cos \theta$. Since the speed is constant, $mg \sin \theta = f_k$.
Thus, the heat generated $Q = W_f = (mg \sin \theta) \cdot s$.
Equating heat generated to heat absorbed: $Q = mc \Delta T$.
$mc \Delta T = (mg \sin \theta) s$.
$\Delta T = \frac{g \sin \theta \cdot s}{c} = \frac{10 \times (3/5) \times 0.8}{420}$.
$\Delta T = \frac{10 \times 0.6 \times 0.8}{420} = \frac{4.8}{420} \approx 0.0114^{\circ} C$.

(B) In a standard phase diagram (Pressure vs Temperature):
$1$. The curve separating the Solid and Liquid phases is the Fusion (or Melting) curve. In the figure,$AO$ separates the Solid and Liquid regions,so $AO$ is the Fusion curve.
$2$. The curve separating the Solid and Vapour phases is the Sublimation curve. In the figure,$BO$ separates the Solid and Vapour regions,so $BO$ is the Sublimation curve.
$3$. The curve separating the Liquid and Vapour phases is the Vaporization (or Boiling) curve. In the figure,$CO$ separates the Liquid and Vapour regions,so $CO$ is the Vaporization curve.
Therefore,$AO =$ Fusion curve,$BO =$ Sublimation curve,and $CO =$ Vaporization curve.

(C) The energy lost by the ball during the bounce is equal to the difference in potential energy at the two heights.
Energy lost $\Delta E = mg(h_1 - h_2)$.
Given $m = 200 \,g = 0.2 \,kg$, $h_1 = 20 \,m$, $h_2 = 10.8 \,m$, $g = 10 \,ms^{-2}$, and $c = 460 \,Jkg^{-1} K^{-1}$.
Assuming the lost energy is absorbed by the ball, $\Delta E = mc \Delta T$.
Equating the two: $mg(h_1 - h_2) = mc \Delta T$.
Canceling $m$ from both sides: $g(h_1 - h_2) = c \Delta T$.
Substituting the values: $10 \times (20 - 10.8) = 460 \times \Delta T$.
$10 \times 9.2 = 460 \times \Delta T$.
$92 = 460 \times \Delta T$.
$\Delta T = \frac{92}{460} = 0.2^{\circ} C$.

(A) The process involves three steps:
$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot S_{ice} \cdot \Delta T = 0.2 \ kg \times 2100 \ J \ kg^{-1} K^{-1} \times 10 \ K = 4200 \ J$
$2$. Melting ice at $0^{\circ}C$ to water at $0^{\circ}C$: $Q_2 = m \cdot L_f = 0.2 \ kg \times 3.35 \times 10^5 \ J \ kg^{-1} = 67000 \ J$
$3$. Heating water from $0^{\circ}C$ to $30^{\circ}C$: $Q_3 = m \cdot S_{water} \cdot \Delta T = 0.2 \ kg \times 4186 \ J \ kg^{-1} K^{-1} \times 30 \ K = 25116 \ J$
Total heat required $Q = Q_1 + Q_2 + Q_3 = 4200 + 67000 + 25116 = 96316 \ J$.

(A) Given: Mass of steam,$m_s = 5 \text{ g}$,Temperature of steam,$T_s = 100^{\circ} \text{C}$.
Mass of ice,$m_i = 5 \text{ g}$,Temperature of ice,$T_i = 0^{\circ} \text{C}$.
Latent heat of vaporization,$L_v = 540 \text{ cal/g}$.
Latent heat of fusion,$L_f = 80 \text{ cal/g}$.
Specific heat of water,$s = 1 \text{ cal/g}^{\circ} \text{C}$.
Step $1$: Calculate heat released by steam to condense into water at $100^{\circ} \text{C}$:
$Q_1 = m_s \times L_v = 5 \times 540 = 2700 \text{ cal}$.
Step $2$: Calculate heat required by ice to melt and reach $100^{\circ} \text{C}$:
Heat to melt ice: $Q_{melt} = m_i \times L_f = 5 \times 80 = 400 \text{ cal}$.
Heat to raise water temperature from $0^{\circ} \text{C}$ to $100^{\circ} \text{C}$: $Q_{rise} = m_i \times s \times \Delta T = 5 \times 1 \times 100 = 500 \text{ cal}$.
Total heat required by ice: $Q_2 = 400 + 500 = 900 \text{ cal}$.
Step $3$: Compare $Q_1$ and $Q_2$:
Since $Q_1 > Q_2$,the steam has more than enough energy to melt the ice and heat the resulting water to $100^{\circ} \text{C}$.
Therefore,the final mixture will be at $100^{\circ} \text{C}$ with some steam remaining.

(D) The density of water is maximum at $4^{\circ} C$.
As the temperature increases above $4^{\circ} C$,the density of water decreases,causing the volume to increase.
Similarly,as the temperature decreases below $4^{\circ} C$,the density of water also decreases due to anomalous expansion,which again causes the volume to increase.
Since the beaker is already full,any increase in volume leads to overflow.
Therefore,water overflows when heated above $4^{\circ} C$ or cooled below $4^{\circ} C$.
Statement $D$ is incorrect because water does overflow when cooled below $4^{\circ} C$.

(B) Given: Height $h = 210 \ m$,Specific heat $c = 4.2 \times 10^3 \ J \ kg^{-1} \ K^{-1}$,Acceleration due to gravity $g = 10 \ m/s^2$.
When water falls,its potential energy is converted into kinetic energy.
Potential Energy $(PE) = mgh$.
According to the problem,$60 \%$ of the kinetic energy is converted into heat energy $(Q)$.
$Q = 0.6 \times PE = 0.6 \times mgh$.
Also,the heat produced is given by $Q = mc\Delta T$,where $\Delta T$ is the rise in temperature.
Equating the two expressions for $Q$:
$mc\Delta T = 0.6 \times mgh$.
Canceling mass $m$ from both sides:
$c\Delta T = 0.6 \times g \times h$.
$\Delta T = \frac{0.6 \times g \times h}{c}$.
Substituting the values:
$\Delta T = \frac{0.6 \times 10 \times 210}{4.2 \times 10^3} = \frac{1260}{4200} = 0.3^{\circ} C$.
Thus,the rise in temperature is $0.3^{\circ} C$.

(B) Given: Mass of hammer $M = 200 \ kg$,mass of steel block $m = 200 \ g = 0.2 \ kg$,velocity of hammer $v = 8 \ ms^{-1}$,and specific heat capacity of steel $s = 460 \ J \ kg^{-1} \ K^{-1}$.
The kinetic energy of the hammer is given by $KE = \frac{1}{2} M v^2$.
Substituting the values: $KE = \frac{1}{2} \times 200 \times 8^2 = 100 \times 64 = 6400 \ J$.
Only $23 \%$ of this energy is utilized to heat the steel block.
Heat energy $H = 23 \% \text{ of } 6400 \ J = \frac{23}{100} \times 6400 = 1472 \ J$.
The heat absorbed by the block is given by $H = m s \Delta T$,where $\Delta T$ is the rise in temperature.
$\Delta T = \frac{H}{m s} = \frac{1472}{0.2 \times 460} = \frac{1472}{92} = 16 \ K$.
Thus,the rise in temperature of the block is $16 \ K$.

(A) . When ice melts into water,the hydrogen-bonded structure of ice collapses,leading to a more compact arrangement of molecules. Thus,the density increases and the volume decreases. $(A-II)$
$B$. When water changes into steam,the molecules move far apart,resulting in a significant increase in volume. $(B-I)$
$C$. Since ice contracts upon melting,according to Clausius-Clapeyron relation,its melting point decreases with an increase in pressure. $(C-IV)$
$D$. Boiling involves a large increase in volume,so increasing the pressure makes it harder for molecules to escape into the vapor phase,thus increasing the boiling point. $(D-III)$
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(B) The potential energy $(PE)$ of the hailstone at height $h$ is given by $PE = mgh$.
Given: $m = 42 \,g = 0.042 \,kg$, $h = 1.8 \,km = 1800 \,m$, $g = 10 \,ms^{-2}$.
$PE = 0.042 \times 10 \times 1800 = 756 \,J$.
This energy is converted into latent heat $(Q)$ to melt a mass $m'$ of the ice: $Q = m' L_{\text{ice}}$.
$756 = m' \times 3.36 \times 10^5$.
$m' = \frac{756}{3.36 \times 10^5} = 225 \times 10^{-5} \,kg = 2.25 \times 10^{-3} \,kg = 2.25 \,g$.
The remaining mass of the hailstone is $m_{remaining} = m - m' = 42 \,g - 2.25 \,g = 39.75 \,g$.

(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.
Heat released by $m \ kg$ of water at $t^{\circ} C$ to become ice at $0^{\circ} C$ is:
$Q_1 = m \times c \times (t - 0) + m \times L_f = m \times 1 \times t + m \times 80 = m(t + 80)$.
Heat absorbed by $(M - m) \ kg$ of water at $t^{\circ} C$ to become steam at $100^{\circ} C$ is:
$Q_2 = (M - m) \times c \times (100 - t) + (M - m) \times L_v = (M - m) \times 1 \times (100 - t) + (M - m) \times 540 = (M - m)(640 - t)$.
Equating $Q_1 = Q_2$:
$m(t + 80) = (M - m)(640 - t)$
$mt + 80m = 640M - Mt - 640m + mt$
$80m + 640m = 640M - Mt$
$720m = M(640 - t)$
$\frac{m}{M} = \frac{640 - t}{720}$.

(A) The kinetic energy of the bullet is converted into heat. Since $\frac{1}{4}$ of the heat is absorbed by the obstacle,$\frac{3}{4}$ of the kinetic energy is used to heat and melt the bullet.
The energy balance equation is: $\frac{3}{4} (\frac{1}{2} m v^2) = m s \Delta \theta + m L$.
Given: $s = 0.03 \ cal \ g^{-1} {}^{\circ} C^{-1} = 0.03 \times 4200 \ J \ kg^{-1} K^{-1} = 126 \ J \ kg^{-1} K^{-1}$.
$\Delta \theta = 600 \ K - 300 \ K = 300 \ K$.
$L = 6 \ cal \ g^{-1} = 6 \times 4200 \ J \ kg^{-1} = 25200 \ J \ kg^{-1}$.
Substituting the values: $\frac{3}{8} v^2 = (126 \times 300) + 25200$.
$\frac{3}{8} v^2 = 37800 + 25200 = 63000$.
$v^2 = \frac{63000 \times 8}{3} = 21000 \times 8 = 168000$.
$v = \sqrt{168000} \approx 409.88 \ ms^{-1} \approx 410 \ ms^{-1}$.

(B) Thermal expansion is the process where the volume of matter changes in response to a change in temperature.
When a substance is heated,its particles gain kinetic energy and move further apart,leading to an increase in volume.
Since density is defined as $\rho = \frac{m}{V}$,where $m$ is mass and $V$ is volume,an increase in volume $V$ for a constant mass $m$ results in a decrease in density $\rho$.
Therefore,expansion during heating decreases the density of the material.

(C) Given: Density of wood at $0^{\circ} C$,$\rho_w = 880 \ kg \ m^{-3}$.
Density of benzene at $0^{\circ} C$,$\rho_b = 900 \ kg \ m^{-3}$.
Coefficient of volume expansion of wood,$\gamma_w = 1.2 \times 10^{-3} \ ^{\circ}C^{-1}$.
Coefficient of volume expansion of benzene,$\gamma_b = 1.5 \times 10^{-3} \ ^{\circ}C^{-1}$.
Initial temperature,$T_1 = 0^{\circ} C$.
Let $T_2$ be the temperature at which the wood just sinks,and $\Delta T = T_2 - T_1$.
The wood sinks when its density equals the density of benzene at temperature $T_2$.
Density at temperature $T$ is given by $\rho_T = \frac{\rho_0}{1 + \gamma \Delta T}$.
Equating densities: $\frac{\rho_w}{1 + \gamma_w \Delta T} = \frac{\rho_b}{1 + \gamma_b \Delta T}$.
Substituting values: $\frac{880}{1 + 1.2 \times 10^{-3} \Delta T} = \frac{900}{1 + 1.5 \times 10^{-3} \Delta T}$.
$880(1 + 1.5 \times 10^{-3} \Delta T) = 900(1 + 1.2 \times 10^{-3} \Delta T)$.
$880 + 1.32 \Delta T = 900 + 1.08 \Delta T$.
$(1.32 - 1.08) \Delta T = 900 - 880$.
$0.24 \Delta T = 20$.
$\Delta T = \frac{20}{0.24} \approx 83.3^{\circ} C$.
Since $T_1 = 0^{\circ} C$,$T_2 = 83.3^{\circ} C$.

(B) The apparent weight $w$ of a body immersed in a liquid is given by $w = V_s(\rho_s - \rho_w)g$,where $V_s$ is the volume of the sphere,$\rho_s$ is the density of the sphere,and $\rho_w$ is the density of water.
Since the volume $V_s$ increases with temperature,we consider the mass $M = V_s \rho_s$ to be constant. Thus,$w = Mg - V_s \rho_w g$.
As temperature increases from $0^{\circ} C$ to $50^{\circ} C$,the volume of the sphere $V_s$ increases,and the density of water $\rho_w$ decreases significantly because the coefficient of cubical expansion of water is greater than that of the metal.
Since $\rho_w$ decreases more rapidly than the volume $V_s$ increases,the buoyant force $F_B = V_s \rho_w g$ decreases as temperature increases.
Therefore,the apparent weight $w = Mg - F_B$ increases with temperature.
Thus,$w_2 > w_1$ or $w_1 < w_2$.

(B) The initial kinetic energy of the pellet is $K = \frac{1}{2} m v^2$.
According to the problem,$50\%$ of this kinetic energy is converted into thermal energy $(Q)$ within the pellet.
Thus,$Q = 0.5 \times K = 0.5 \times \frac{1}{2} m v^2 = \frac{1}{4} m v^2$.
The thermal energy gained by the pellet is also given by the formula $Q = m c \Delta T$,where $\Delta T$ is the rise in temperature.
Equating the two expressions for $Q$: $m c \Delta T = \frac{1}{4} m v^2$.
Solving for $\Delta T$: $\Delta T = \frac{m v^2}{4 m c} = \frac{v^2}{4 c}$.

(A) Mass of ice $m = 8 \ g = 0.008 \ kg$.
Step $1$: Heat to raise ice from $-20^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot c_{ice} \cdot \Delta T = 0.008 \times 2100 \times 20 = 336 \ J$.
Step $2$: Heat to melt ice at $0^{\circ}C$ to water at $0^{\circ}C$: $Q_2 = m \cdot L_f = 0.008 \times 336 \times 10^3 = 2688 \ J$.
Step $3$: Heat to raise water from $0^{\circ}C$ to $100^{\circ}C$: $Q_3 = m \cdot c_{water} \cdot \Delta T = 0.008 \times 4200 \times 100 = 3360 \ J$.
Step $4$: Heat to convert water at $100^{\circ}C$ to steam at $100^{\circ}C$: $Q_4 = m \cdot L_v = 0.008 \times 2.268 \times 10^6 = 18144 \ J$.
Total heat $Q = Q_1 + Q_2 + Q_3 + Q_4 = 336 + 2688 + 3360 + 18144 = 24528 \ J \approx 24.5 \ kJ$.

(B) Given: $m_{\text{ice}} = 54 \ g$,$T_{\text{ice}} = -20^{\circ} C$,$m_{\text{steam}} = 25 \ g$,$T_{\text{steam}} = 100^{\circ} C$.
Heat required to bring ice to $0^{\circ} C$: $Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T = 54 \times 2.1 \times 20 = 2268 \ J$.
Heat required to melt the ice at $0^{\circ} C$: $Q_2 = m_{\text{ice}} \cdot L_f = 54 \times 334 = 18036 \ J$.
Total heat required to convert ice at $-20^{\circ} C$ to water at $0^{\circ} C$: $Q_{\text{total}} = Q_1 + Q_2 = 2268 + 18036 = 20304 \ J$.
Heat released by condensing $25 \ g$ of steam at $100^{\circ} C$ to water at $100^{\circ} C$: $Q_{\text{condense}} = m_{\text{steam}} \cdot L_v = 25 \times 2260 = 56500 \ J$.
Since $Q_{\text{condense}} > Q_{\text{total}}$,all ice melts and the water temperature rises to $100^{\circ} C$.
Heat required to raise the temperature of $54 \ g$ of water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m_{\text{ice}} \cdot c_{\text{water}} \cdot \Delta T = 54 \times 4.2 \times 100 = 22680 \ J$.
Remaining heat available for condensation: $Q_{\text{rem}} = Q_{\text{condense}} - (Q_{\text{total}} + Q_3) = 56500 - (20304 + 22680) = 56500 - 42984 = 13516 \ J$.
Mass of steam condensed by this remaining heat: $m_{\text{condensed}} = Q_{\text{rem}} / L_v = 13516 / 2260 \approx 6 \ g$.
Final mass of water = $54 \ g$ (melted ice) + $(25 - 6) \ g$ (condensed steam) = $73 \ g$.
Final mass of steam = $6 \ g$ (remaining steam).
Thus,the mixture contains $73 \ g$ of water and $6 \ g$ of steam at $100^{\circ} C$.

(C) Let $c$ be the specific heat capacity of the material. The heat supplied is given by the formula $Q = mc\Delta T$.
For the object of mass $2m$,the initial temperature is $T$ and the final temperature is $2T$. The change in temperature is $\Delta T_1 = 2T - T = T$.
Thus,the heat supplied is $Q = (2m)c(T) = 2mcT$.
For the object of mass $m$,the initial temperature is $2T$. Let the final temperature be $T_f$. The change in temperature is $\Delta T_2 = T_f - 2T$.
Since the same heat $Q$ is supplied,we have $Q = m c (T_f - 2T)$.
Equating the two expressions for $Q$: $2mcT = mc(T_f - 2T)$.
Dividing both sides by $mc$: $2T = T_f - 2T$.
Solving for $T_f$: $T_f = 2T + 2T = 4T$.

(A) Mass of ice,$m = 10 \,kg = 10^4 \,g$.
Specific heat capacity of ice,$c = 0.5 \,cal \,g^{-1} \,^{\circ}C^{-1} = 2093.4 \,J \,kg^{-1} \,K^{-1}$.
Latent heat of fusion of ice,$L = 80 \,cal \,g^{-1} = 334944 \,J \,kg^{-1}$.
The total heat energy $Q$ required is the sum of heat to raise the temperature of ice to $0^{\circ}C$ and the heat to melt the ice at $0^{\circ}C$.
$Q = m c \Delta T + m L$
$Q = 10 \,kg \times 2093.4 \,J \,kg^{-1} \,K^{-1} \times (0 - (-10)) \,K + 10 \,kg \times 334944 \,J \,kg^{-1}$
$Q = 10 \times 2093.4 \times 10 + 3349440$
$Q = 209340 + 3349440$
$Q = 3558780 \,J \approx 357 \times 10^4 \,J$.

(C) The heat energy required to raise the temperature of a substance is given by $\Delta Q = m c \Delta T$.
Since mass $m = d V$ (where $d$ is density and $V$ is volume),we have $\Delta Q = d V c \Delta T$.
The heat energy required per unit volume is $\frac{\Delta Q}{V} = d c \Delta T$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{5}{6}$ and the ratio of specific heat capacities $\frac{c_1}{c_2} = \frac{3}{5}$.
For the same temperature rise,$\Delta T_1 = \Delta T_2 = \Delta T$.
The ratio of heat energies required per unit volume is $\frac{(\Delta Q/V)_1}{(\Delta Q/V)_2} = \frac{d_1 c_1 \Delta T}{d_2 c_2 \Delta T} = \frac{d_1}{d_2} \times \frac{c_1}{c_2}$.
Substituting the given values: $\frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2}$.

(C) The potential energy lost by the water in falling from a height $h$ is given by $\Delta U = mgh$.
According to the problem,$10 \%$ of this energy is converted into heat $(\Delta Q)$ to raise the temperature of the water.
Thus,$\Delta Q = 0.1 \times \Delta U$.
Substituting the formulas for heat and potential energy: $ms \Delta T = 0.1 \times mgh$.
Canceling mass $m$ from both sides: $s \Delta T = 0.1 \times gh$.
Rearranging for $\Delta T$: $\Delta T = \frac{0.1 \times g \times h}{s}$.
Substituting the given values: $\Delta T = \frac{0.1 \times 10 \times 20}{4000}$.
$\Delta T = \frac{20}{4000} = \frac{1}{200} = 0.005^{\circ} C$.

(A) The process involves several steps to convert ice at $-10^{\circ} C$ to steam at $110^{\circ} C$:
$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: $Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T = 0.04 \ kg \times 0.5 \ kcal/kg^{\circ} C \times 10^{\circ} C = 0.2 \ kcal$.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: $Q_2 = m \cdot L_{\text{fusion}} = 0.04 \ kg \times 80 \ kcal/kg = 3.2 \ kcal$.
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m \cdot c_{\text{water}} \cdot \Delta T = 0.04 \ kg \times 1 \ kcal/kg^{\circ} C \times 100^{\circ} C = 4.0 \ kcal$.
$4$. Vaporizing water at $100^{\circ} C$ to steam at $100^{\circ} C$: $Q_4 = m \cdot L_{\text{vap}} = 0.04 \ kg \times 540 \ kcal/kg = 21.6 \ kcal$.
$5$. Heating steam from $100^{\circ} C$ to $110^{\circ} C$: $Q_5 = m \cdot c_{\text{steam}} \cdot \Delta T = 0.04 \ kg \times 0.48 \ kcal/kg^{\circ} C \times 10^{\circ} C = 0.192 \ kcal$.
Total thermal energy $Q = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 0.2 + 3.2 + 4.0 + 21.6 + 0.192 = 29.192 \ kcal$.

(C) Let the mass of liquid ammonia be $m$ and the rate of heat supplied by the heater be $r$.
According to the principle of calorimetry,the heat supplied to raise the temperature is given by $Q_1 = r \times 14 = m \times c \times \Delta T$.
Substituting the values: $r \times 14 = m \times 4.9 \times (50 - 15)$.
$r \times 14 = m \times 4.9 \times 35$ --- (Equation $1$).
The heat supplied to convert the liquid to vapour at the boiling point is $Q_2 = r \times 92 = m \times L$.
$r \times 92 = m \times L$ --- (Equation $2$).
Dividing Equation $2$ by Equation $1$:
$\frac{r \times 92}{r \times 14} = \frac{m \times L}{m \times 4.9 \times 35}$.
$\frac{92}{14} = \frac{L}{4.9 \times 35}$.
$L = \frac{92 \times 4.9 \times 35}{14}$.
$L = 92 \times 4.9 \times 2.5 = 1127 \text{ kJ/kg}$.

(A) The time period of a pendulum clock is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
The fractional change in time period due to a change in temperature $\Delta \theta$ is given by $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta$,where $\alpha$ is the coefficient of linear expansion.
Given: $\alpha = 12 \times 10^{-6} /{ }^{\circ} C$,$\Delta \theta = 47^{\circ} C - 32^{\circ} C = 15^{\circ} C$.
$\frac{\Delta T}{T} = \frac{1}{2} \times 12 \times 10^{-6} \times 15 = 90 \times 10^{-6}$.
The time lost or gained in a day $(86400 \ s)$ is $\Delta t = \frac{\Delta T}{T} \times 86400$.
$\Delta t = 90 \times 10^{-6} \times 86400 = 7.776 \ s \approx 7.8 \ s$.
Since the temperature increases,the length of the pendulum increases,the time period increases,and the clock runs slow.

(D) Statement $(I)$ is true: $A$ calorimeter is indeed a device used for measuring the heat involved in physical or chemical processes.
Statement $(II)$ is false: While skating,the pressure applied by the skates on the ice lowers the melting point of the ice (Regelation),causing it to melt into a thin layer of water that acts as a lubricant. The statement incorrectly attributes this to an increase in temperature.
Statement $(III)$ is true: In an isolated system like a calorimeter,the heat lost by the hot body equals the heat gained by the cold body,meaning the total internal energy of the system is conserved.
Therefore,statements $(I)$ and $(III)$ are true,while statement $(II)$ is false.

(C) $1$. Heat required to melt $100 \ g$ of ice at $0^{\circ} C$ to water at $0^{\circ} C$: $Q_1 = m \times L_f = 100 \ g \times 80 \ cal/g = 8000 \ cal$.
$2$. Heat required to raise the temperature of $100 \ g$ of water from $0^{\circ} C$ to $100^{\circ} C$: $Q_2 = m \times s \times \Delta T = 100 \ g \times 1 \ cal/g^{\circ} C \times 100^{\circ} C = 10000 \ cal$.
$3$. Total heat used so far: $Q_{total} = Q_1 + Q_2 = 8000 + 10000 = 18000 \ cal$.
$4$. Remaining heat: $Q_{rem} = 22320 \ cal - 18000 \ cal = 4320 \ cal$.
$5$. This remaining heat will convert some water at $100^{\circ} C$ into steam: $m_{steam} = Q_{rem} / L_v = 4320 \ cal / 540 \ cal/g = 8 \ g$.
$6$. Final amount of water remaining: $100 \ g - 8 \ g = 92 \ g$ of water at $100^{\circ} C$.

(A) The process of heating ice from $-20^{\circ} C$ to steam at $100^{\circ} C$ involves several stages:
$1$. Heating ice from $-20^{\circ} C$ to $0^{\circ} C$: The temperature increases linearly with heat supplied.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: The temperature remains constant (phase change).
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: The temperature increases linearly with heat supplied.
$4$. Boiling water at $100^{\circ} C$ to steam at $100^{\circ} C$: The temperature remains constant (phase change).
Therefore,the temperature-heat graph should show two linear increases and two horizontal constant-temperature segments. Option $A$ correctly represents this behavior.

(C) The potential energy of the water at the top is converted into heat energy at the bottom.
Using the principle of conservation of energy: $mgh = J \cdot ms \Delta t$,where $J$ is the mechanical equivalent of heat ($J = 4.2 \ J/cal$ or $4186 \ J/kg^{\circ}C$).
Here,$s$ is the specific heat capacity of water,$s = 4186 \ J/kg^{\circ}C$.
Rearranging for $\Delta t$: $\Delta t = \frac{gh}{s}$.
Substituting the values: $\Delta t = \frac{9.8 \times 50}{4186} \approx 0.117^{\circ} C$.

(B) When heating water from $-20^{\circ}C$ to $120^{\circ}C$,the following phase changes occur:
$1$. From $-20^{\circ}C$ to $0^{\circ}C$,ice is heated (temperature increases).
$2$. At $0^{\circ}C$,ice melts into water (phase change,temperature remains constant,creating a plateau).
$3$. From $0^{\circ}C$ to $100^{\circ}C$,water is heated (temperature increases).
$4$. At $100^{\circ}C$,water boils into steam (phase change,temperature remains constant,creating a second plateau).
$5$. From $100^{\circ}C$ to $120^{\circ}C$,steam is heated (temperature increases).
Therefore,the graph must show two horizontal plateaus at $0^{\circ}C$ and $100^{\circ}C$. This corresponds to the graph shown in option $B$.

(C) Step $1$: Calculate the heat released by $x \ \text{g}$ of water cooling from $50^\circ \text{C}$ to $0^\circ \text{C}$.
$Q_1 = m c \Delta T = (x \times 10^{-3} \ \text{kg}) \times 4200 \ \text{J/kg}\cdot \text{K} \times (50 - 0) \ \text{K} = 210x \ \text{J}$.
Step $2$: Calculate the heat required to evaporate $(1000 - x) \ \text{g}$ of water initially at $50^\circ \text{C}$. This involves heating the water to $100^\circ \text{C}$ and then vaporizing it.
$Q_2 = m' c \Delta T' + m' L = [(1000 - x) \times 10^{-3} \ \text{kg}] \times [4200 \ \text{J/kg}\cdot \text{K} \times (100 - 50) \ \text{K} + 2256000 \ \text{J/kg}]$.
$Q_2 = (1000 - x) \times 10^{-3} \times [210000 + 2256000] = (1000 - x) \times 10^{-3} \times 2466000 = 2466(1000 - x) \ \text{J}$.
Step $3$: Equate $Q_1$ and $Q_2$ to find $x$.
$210x = 2466(1000 - x) \implies 210x = 2466000 - 2466x$.
$2676x = 2466000 \implies x = \frac{2466000}{2676} \approx 921.52$.
The closest integer value is $922$.