Heat Capacity, Specific Heat and Molar Specific Heat Questions in English

(C) The formula for heat exchange is given by $Q = m \cdot c \cdot \Delta \theta$,where $Q$ is the heat,$m$ is the mass,$c$ is the specific heat,and $\Delta \theta$ is the change in temperature.
Rearranging for specific heat,we get $c = \frac{Q}{m \cdot \Delta \theta}$.
If the specific heat $c$ is infinite $(c = \infty)$,then the denominator $m \cdot \Delta \theta$ must be zero.
Since mass $m$ cannot be zero,it implies that $\Delta \theta = 0$.
This means that there is no change in temperature regardless of whether heat is absorbed or released by the substance,which is characteristic of a phase change process.

(C) The formula for heat lost or gained without a change of state is given by $\Delta Q = mc \Delta \theta$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta \theta$ is the change in temperature.
Since the formula depends on mass $(m)$,specific heat $(c)$,and temperature change $(\Delta \theta)$,the factor 'Relative density' is not required for this calculation.
Therefore,the correct option is $C$.

(D) Water is used as a coolant in engine radiators primarily because it has a very high specific heat capacity $(c \approx 4186 \ J/kg \cdot K)$.
This property allows water to absorb a large amount of heat energy from the engine with a relatively small rise in its own temperature.
Consequently,it is highly efficient at transferring heat away from the engine components to the radiator,where it can be dissipated into the atmosphere.

(A) The formula for heat energy gained is $Q = m \cdot c \cdot \Delta \theta$.
Given:
Mass of water,$m = 5\, kg$.
Specific heat of water,$c = 4.2\, kJ\, kg^{-1} {^{\circ}C}^{-1}$.
Initial temperature,$\theta_1 = 20^{\circ}C$.
Final temperature (boiling point),$\theta_2 = 100^{\circ}C$.
Change in temperature,$\Delta \theta = \theta_2 - \theta_1 = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C$.
Substituting the values into the formula:
$Q = 5\, kg \times 4.2\, kJ\, kg^{-1} {^{\circ}C}^{-1} \times 80^{\circ}C$.
$Q = 5 \times 4.2 \times 80\, kJ$.
$Q = 21 \times 80\, kJ$.
$Q = 1680\, kJ$.

(B) Thermal capacity is defined as the product of mass $(m)$ and specific heat capacity $(c)$.
Thermal Capacity $= m \times c = (V \times \rho) \times c$,where $V$ is volume,$\rho$ is density,and $c$ is specific heat.
Since both spheres are made of the same substance,their density $(\rho)$ and specific heat $(c)$ are identical.
Therefore,the ratio of thermal capacities is equal to the ratio of their volumes.
Volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
Given the ratio of diameters is $1:2$,the ratio of radii $(r_1:r_2)$ is also $1:2$.
Ratio of thermal capacities $= \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Thus,the ratio is $1:8$.

(B) The heat energy $Q$ required to change the temperature of a body of mass $m$ and specific heat $c$ by an amount $\Delta \theta$ is given by the formula: $Q = m \cdot c \cdot \Delta \theta$.
If we set the change in temperature $\Delta \theta = 1^oC$ (or $1\,K$),the equation becomes $Q = m \cdot c$.
The product $m \cdot c$ is defined as the thermal capacity (or heat capacity) of the body.
Therefore,the amount of heat required to raise the temperature of a body by $1^oC$ is its thermal capacity.

(D) Thermal capacity is defined as the product of the mass of the substance $(m)$ and its specific heat capacity $(c)$.
Given:
Mass $(m)$ $= 40\, g$
Specific heat $(c)$ $= 0.2\, cal/g/^{\circ}C$
Thermal capacity $= m \times c$
Thermal capacity $= 40\, g \times 0.2\, cal/g/^{\circ}C = 8\, cal/^{\circ}C$.
Therefore, the correct option is $D$.

(B) The formula for heat energy is $Q = m \cdot c \cdot \Delta \theta$, where $c$ is the specific heat capacity.
Rearranging for $c$, we get $c = \frac{Q}{m \cdot \Delta \theta}$.
Here, $Q$ is the heat energy (a constant) and $m$ is the mass (a constant).
Therefore, $c \propto \frac{1}{\Delta \theta}$.
Since the size of one degree on the Fahrenheit scale is smaller than the size of one degree on the Celsius scale $(1^\circ C = 1.8^\circ F)$, a temperature interval $\Delta \theta$ expressed in $^\circ F$ will have a larger numerical value than the same interval expressed in $^\circ C$.
Because the denominator $\Delta \theta$ increases when changing from $^\circ C$ to $^\circ F$, the numerical value of specific heat $c$ must decrease.

(B) The heat energy $Q$ required to raise the temperature of a mass $m$ of water by $\Delta T$ is given by the formula $Q = mc\Delta T$.
Here,the specific heat capacity of water $c = 4200\, J/(kg \cdot ^{\circ}C)$ or $4.2\, J/(g \cdot ^{\circ}C)$.
For option $(B)$: $m = 100\, g = 0.1\, kg$ and $\Delta T = 10^{\circ}C$.
$Q = 0.1\, kg \times 4200\, J/(kg \cdot ^{\circ}C) \times 10^{\circ}C = 4200\, J$.
Thus,$4200\, J$ of energy is required to increase the temperature of $100\, g$ of water by $10^{\circ}C$.

(A) The formula for specific heat capacity $(c)$ is given by $c = \frac{Q}{m \cdot \Delta \theta}$,where $Q$ is the heat energy,$m$ is the mass,and $\Delta \theta$ is the change in temperature.
Substituting the $SI$ units: $Q$ is measured in Joules $(J)$,$m$ is measured in kilograms $(kg)$,and $\Delta \theta$ is measured in degrees Celsius $(^\circ C)$ or Kelvin $(K)$.
Therefore,the unit of specific heat is $\frac{J}{kg \cdot ^\circ C}$ or $J/kg\,^\circ C$.

(A) The specific heat capacity is defined as the amount of heat required to raise the temperature of $1 \ kg$ of a substance by $1 \ K$.
Among the given options,water has the highest specific heat capacity,which is approximately $4186 \ J/(kg \cdot K)$.
Alcohol,glycerine,and oil have significantly lower specific heat capacities compared to water.
Therefore,the correct option is $A$.

(C) For a process at constant volume,the heat supplied $(\Delta Q)_V$ is equal to the change in internal energy $(\Delta U)$.
Given: $\Delta U = 80 \ J$,$\Delta T = 120^{\circ}C - 100^{\circ}C = 20 \ K$.
The total heat capacity at constant volume $(C_{total})$ is defined as $C_{total} = \frac{(\Delta Q)_V}{\Delta T}$.
Since $(\Delta Q)_V = \Delta U$,we have $C_{total} = \frac{\Delta U}{\Delta T}$.
Substituting the values: $C_{total} = \frac{80 \ J}{20 \ K} = 4 \ J/K$.

(D) Water is used to cool the radiators of engines because water has a very high specific heat capacity $(c = 4186 \ J/kg \cdot K)$.
Due to this high specific heat,water can absorb a large amount of heat energy from the engine with a relatively small increase in its own temperature.
This makes it an extremely efficient coolant for maintaining the operating temperature of an engine.

(B) The thermal capacity (or heat capacity) of a body is defined as the amount of heat required to raise the temperature of the entire body by $1^{\circ}C$ (or $1 \ K$).
Mathematically,$C = m \times s$,where $m$ is the mass of the body and $s$ is the specific heat capacity of the material.
Therefore,the correct option is $B$.

(B) The heat required to change the temperature of a body is given by the formula $Q = m \cdot c \cdot \Delta \theta$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta \theta$ is the change in temperature.
If the change in temperature $\Delta \theta = 1 ^\circ C$ (or $1 \, K$),then the heat required is $Q = m \cdot c$.
This product $m \cdot c$ is defined as the thermal capacity (or heat capacity) of the body.

(B) Heat capacity $H_C$ is defined as the product of mass $m$ and specific heat capacity $c$,i.e.,$H_C = m \cdot c$.
Since both spheres are made of the same material,their specific heat capacities $c$ are equal.
Therefore,the heat capacity is directly proportional to the mass: $H_C \propto m$.
Since $m = V \cdot \rho$ (where $V$ is volume and $\rho$ is density),and density $\rho$ is the same for both,we have $H_C \propto V$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$,so $H_C \propto r^3$.
Given the ratio of diameters $d_1 : d_2 = 1 : 2$,the ratio of radii $r_1 : r_2$ is also $1 : 2$.
Thus,the ratio of heat capacities is $\frac{(H_C)_1}{(H_C)_2} = \left( \frac{r_1}{r_2} \right)^3 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.

(B) The rate of heat loss is given by $P = \frac{dQ}{dt}$.
We know that $dQ = C \cdot dT$,where $C$ is the heat capacity and $dT$ is the change in temperature.
Therefore,the rate of heat loss is $P = C \cdot \frac{dT}{dt}$.
Given that the rate of heat loss $P$ is the same for both bodies,we have $C_1 \cdot \left(\frac{dT}{dt}\right)_1 = C_2 \cdot \left(\frac{dT}{dt}\right)_2$.
This implies $\frac{(\frac{dT}{dt})_1}{(\frac{dT}{dt})_2} = \frac{C_2}{C_1}$.
Given the ratio of heat capacities $\frac{C_1}{C_2} = \frac{1}{4}$,we find $\frac{C_2}{C_1} = \frac{4}{1}$.
Thus,the ratio of the rate of fall of temperatures is $4:1$.

(B) The heat required to change the temperature of a substance by $dT$ is given by $dQ = m C dT$.
Substituting the given expression for specific heat $C = \alpha T^2 + \beta T + \gamma$,we get:
$dQ = m (\alpha T^2 + \beta T + \gamma) dT$
To find the total heat $Q$ required to raise the temperature from $0^{\circ}C$ to $T_o^{\circ}C$,we integrate the expression:
$Q = \int_{0}^{T_o} m (\alpha T^2 + \beta T + \gamma) dT$
$Q = m \left[ \frac{\alpha T^3}{3} + \frac{\beta T^2}{2} + \gamma T \right]_{0}^{T_o}$
$Q = m \left( \frac{\alpha T_o^3}{3} + \frac{\beta T_o^2}{2} + \gamma T_o \right)$
$Q = \frac{m \alpha T_o^3}{3} + \frac{m \beta T_o^2}{2} + m \gamma T_o$

(C) The heat supplied $Q$ is given by $Q = mc\Delta T$,where $m$ is the mass,$c$ is the specific heat,and $\Delta T$ is the change in temperature.
Assuming the rate of heat supply $P = Q/t$ is constant,we have $P \cdot t = mc\Delta T$.
Therefore,$c = \frac{P \cdot t}{m \cdot \Delta T}$.
For a given mass $m$ and a fixed temperature change $\Delta T$,the specific heat $c$ is directly proportional to the time $t$ taken $(c \propto t)$.
From the graph,for a given temperature increase $\Delta T$,the substance $C$ takes the maximum time $t$.
Thus,substance $C$ has the highest specific heat.

(C) The specific heat capacity $C$ is defined as $C = \frac{dQ}{m \cdot dT}$.
If the specific heat $C$ is infinite,then $dT$ must be zero for any non-zero heat exchange $dQ$.
This implies that the temperature of the substance remains constant even when heat is added or removed.
This is characteristic of a phase change (e.g.,melting or boiling) where the substance absorbs or releases latent heat without a change in temperature.

(B) Given: Molar mass $M_w = 50 \, g/mol$,mass of sample $m = 25 \, g$,heat supplied $Q = 300 \, J$,change in temperature $\Delta \theta = 50^{\circ}C - 30^{\circ}C = 20^{\circ}C$.
First,calculate the specific heat capacity $(S)$:
$Q = m \cdot S \cdot \Delta \theta$
$300 = 25 \cdot S \cdot 20$
$300 = 500 \cdot S$
$S = \frac{300}{500} = 0.6 \, J/g \cdot ^{\circ}C$.
Now,calculate the molar heat capacity $(C_m)$:
$C_m = M_w \cdot S$
$C_m = 50 \, g/mol \cdot 0.6 \, J/g \cdot ^{\circ}C$
$C_m = 30 \, J/mol \cdot ^{\circ}C$.
Thus,the correct option is $B$.

(A) The mass of water is $m = 5 \, kg$.
Initial temperature $T_i = 20^oC$.
Final temperature (boiling point) $T_f = 100^oC$.
Specific heat capacity of water $c = 4.2 \, kJ \, kg^{-1} \, ^oC^{-1}$.
The heat required $Q$ is given by the formula $Q = mc\Delta T$.
Change in temperature $\Delta T = T_f - T_i = 100^oC - 20^oC = 80^oC$.
Substituting the values: $Q = 5 \, kg \times 4.2 \, kJ \, kg^{-1} \, ^oC^{-1} \times 80^oC$.
$Q = 5 \times 4.2 \times 80 = 1680 \, kJ$.

(A) Given: Mass $m = 25 \, g = 0.025 \, kg$,Heat $Q = 300 \, J$,Change in temperature $\Delta T = 50^{\circ}C - 30^{\circ}C = 20^{\circ}C$.
First,calculate the specific heat capacity $(s)$:
$Q = ms\Delta T \implies s = \frac{Q}{m\Delta T} = \frac{300 \, J}{0.025 \, kg \times 20^{\circ}C} = \frac{300}{0.5} = 600 \, J/kg^{\circ}C$.
Next,calculate the heat capacity $(C)$:
$C = ms = 0.025 \, kg \times 600 \, J/kg^{\circ}C = 15 \, J/^{\circ}C$.
Thus,the heat capacity is $15 \, J/^{\circ}C$ and the specific heat capacity is $600 \, J/kg^{\circ}C$.

(B) The heat capacity $C$ of an object is given by $C = mc$,where $m$ is the mass and $c$ is the specific heat capacity.
Since both spheres are made of the same material,their specific heat capacity $c$ is the same.
Mass $m$ can be expressed as $m = V \rho$,where $V$ is the volume and $\rho$ is the density.
For a sphere,$V = \frac{4}{3} \pi r^3$.
Therefore,the ratio of heat capacities is $\frac{C_1}{C_2} = \frac{m_1 c}{m_2 c} = \frac{m_1}{m_2} = \frac{V_1 \rho}{V_2 \rho} = \frac{V_1}{V_2}$.
Substituting the volume formula,$\frac{C_1}{C_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \left( \frac{r_1}{r_2} \right)^3$.
Given the ratio of diameters is $1:2$,the ratio of radii is also $r_1:r_2 = 1:2$.
Thus,$\frac{C_1}{C_2} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
The ratio of their heat capacities is $1:8$.

(A) Given: Mass $m = 25 \, g = 0.025 \, kg$,Heat $Q = 300 \, J$,Change in temperature $\Delta T = 45^{\circ}C - 25^{\circ}C = 20^{\circ}C$.
Heat capacity $(C)$ is defined as $C = \frac{Q}{\Delta T} = \frac{300}{20} = 15 \, J/^{\circ}C$.
Specific heat capacity $(c)$ is defined as $c = \frac{C}{m} = \frac{15}{0.025} = 600 \, J/kg \cdot ^{\circ}C$.
Therefore,the heat capacity is $15 \, J/^{\circ}C$ and the specific heat capacity is $600 \, J/kg \cdot ^{\circ}C$.

(C) The heat required $Q$ is given by the integral $Q = \int m c \, dt$.
Given $m = 2 \, g$,$c = 0.20 + 0.14t + 0.023t^2$,and temperature limits $t_1 = 5 \, ^\circ C$ to $t_2 = 15 \, ^\circ C$.
$Q = \int_{5}^{15} 2 \times (0.20 + 0.14t + 0.023t^2) \, dt$
$Q = 2 \times [0.20t + 0.07t^2 + \frac{0.023}{3}t^3]_{5}^{15}$
$Q = 2 \times [(0.20(15) + 0.07(15^2) + \frac{0.023}{3}(15^3)) - (0.20(5) + 0.07(5^2) + \frac{0.023}{3}(5^3))]$
$Q = 2 \times [(3 + 15.75 + 25.875) - (1 + 1.75 + 0.9583)]$
$Q = 2 \times [44.625 - 3.7083] = 2 \times 40.9167 \approx 81.83 \, cal$.
Rounding to the nearest integer,we get $82 \, cal$.

(B) The heat energy required to raise the temperature of water is given by $Q = mc\Delta\theta$.
Here,mass $m = 1\, kg$ (since $1\, litre$ of water has a mass of $1\, kg$),specific heat capacity $c = 4180\, J/(kg\cdot ^oC)$,and change in temperature $\Delta\theta = 40\, ^oC - 20\, ^oC = 20\, ^oC$.
So,$Q = 1 \times 4180 \times 20 = 83600\, J$.
The power of the heater is $P = 836\, W$.
The time $t$ required is given by $t = \frac{Q}{P}$.
$t = \frac{83600}{836} = 100\, sec$.

(B) First,convert the temperatures from Fahrenheit to Celsius using the formula $C = (F - 32) \times \frac{5}{9}$.
Normal temperature: $98.6^\circ F = (98.6 - 32) \times \frac{5}{9} = 66.6 \times \frac{5}{9} = 37^\circ C$.
Fever temperature: $102.2^\circ F = (102.2 - 32) \times \frac{5}{9} = 70.2 \times \frac{5}{9} = 39^\circ C$.
The change in temperature is $\Delta \theta = 39^\circ C - 37^\circ C = 2^\circ C$.
The mass of the man is $m = 80 \, kg = 80,000 \, g$.
The specific heat capacity of water is $s = 1 \, cal/g^\circ C$.
The heat required is given by $\Delta Q = m \cdot s \cdot \Delta \theta$.
$\Delta Q = 80,000 \, g \times 1 \, cal/g^\circ C \times 2^\circ C = 160,000 \, cal$.
Since $1,000 \, cal = 1 \, kcal$,we have $\Delta Q = 160 \, kcal$.

(B) The heat energy $dQ$ required to raise the temperature of a unit mass $(m = 1)$ by a small amount $dT$ is given by $dQ = m \cdot S \cdot dT$.
Substituting $S = aT^3$ and $m = 1$,we get $dQ = aT^3 \ dT$.
To find the total heat energy $Q$ required to raise the temperature from $T_1 = 1 \ K$ to $T_2 = 2 \ K$,we integrate the expression:
$Q = \int_{1}^{2} aT^3 \ dT$
$Q = a \left[ \frac{T^4}{4} \right]_{1}^{2}$
$Q = \frac{a}{4} (2^4 - 1^4)$
$Q = \frac{a}{4} (16 - 1)$
$Q = \frac{15 \ a}{4}$

(D) Heat capacity $(C)$ is defined as the product of mass $(m)$ and specific heat capacity $(c)$,i.e.,$C = mc$.
Mass is given by the product of density $(\rho)$ and volume $(V)$,i.e.,$m = \rho V$.
For material $A$: Density $\rho_A = 1500 \ kg/m^3$,Volume $V_A = 8 \ units$. Mass $m_A = 1500 \times 8 = 12000 \ units$.
For material $B$: Density $\rho_B = 2000 \ kg/m^3$,Volume $V_B = 12 \ units$. Mass $m_B = 2000 \times 12 = 24000 \ units$.
Given that the heat capacity of $8$ volumes of $A$ is equal to the heat capacity of $12$ volumes of $B$:
$m_A \times c_A = m_B \times c_B$
$12000 \times c_A = 24000 \times c_B$
$c_A = 2 c_B$
Therefore,the ratio of specific heats $c_A : c_B = 2 : 1$.

(A) The change in internal energy $\Delta U$ for a liquid undergoing heating,while ignoring expansion,is equal to the heat supplied $\Delta Q$.
Given:
Mass $m = 100 \ g = 0.1 \ kg$
Specific heat $c = 4184 \ J/kg/K$
Change in temperature $\Delta T = 50^\circ C - 30^\circ C = 20 \ K$
Using the formula $\Delta U = mc\Delta T$:
$\Delta U = 0.1 \ kg \times 4184 \ J/kg/K \times 20 \ K$
$\Delta U = 8368 \ J$
Converting to $kJ$:
$\Delta U = 8.368 \ kJ \approx 8.4 \ kJ$.

(C) The rate of heat supply is constant, so $\frac{dQ}{dt} = P$ (where $P$ is a constant).
By definition, $C = \frac{dQ}{dT}$, so $dQ = C \cdot dT = aT^3 \cdot dT$.
Substituting this into the rate equation: $\frac{dQ}{dt} = aT^3 \frac{dT}{dt} = P$.
Rearranging gives $T^3 \cdot dT = \frac{P}{a} \cdot dt$.
Integrating both sides: $\int T^3 \cdot dT = \int \frac{P}{a} \cdot dt$.
This yields $\frac{T^4}{4} = \frac{P}{a} \cdot t + \text{constant}$.
Assuming $T = 0$ at $t = 0$, the constant is $0$, so $T^4 = \frac{4P}{a} \cdot t$.
Thus, $T = \left( \frac{4P}{a} \right)^{1/4} \cdot t^{1/4}$.
Since $T \propto t^{1/4}$, the graph of $T$ versus $t$ will be a curve that starts at the origin and has a decreasing slope as $t$ increases (concave down), which corresponds to Graph $C$.

(B) The heat capacity of a body is defined as the amount of heat required to raise the temperature of the entire body by $1\ ^oC$.
Mathematically,it is given by $C = \frac{Q}{\Delta T}$,where $Q$ is the heat supplied and $\Delta T$ is the change in temperature.
Since the question asks for the heat required to raise the temperature of the body (not per unit mass) by $1\ ^oC$,the correct term is heat capacity.

(B) The rate of cooling is given by $P = \frac{dQ}{dt} = mS \left| \frac{dT}{dt} \right|$,where $P$ is constant power,$m$ is mass,and $S$ is specific heat capacity.
Thus,$S = \frac{P}{m} \left| \frac{dt}{dT} \right|$. Since $P$ and $m$ are constant,$S \propto \left| \frac{dt}{dT} \right|$.
For the gas phase (from $t=0$ to $t=10$ s,$T=100^{\circ}C$ to $70^{\circ}C$): $\Delta t = 10$ s,$\Delta T = 30^{\circ}C$. So,$S_{gas} \propto \frac{10}{30} = \frac{1}{3}$.
For the liquid phase (from $t=20$ to $t=40$ s,$T=70^{\circ}C$ to $50^{\circ}C$): $\Delta t = 20$ s,$\Delta T = 20^{\circ}C$. So,$S_{liquid} \propto \frac{20}{20} = 1$.
For the solid phase (from $t=80$ to $t=100$ s,$T=50^{\circ}C$ to $10^{\circ}C$): $\Delta t = 20$ s,$\Delta T = 40^{\circ}C$. So,$S_{solid} \propto \frac{20}{40} = \frac{1}{2}$.
Now,$S_{solid} : S_{liquid} : S_{gas} = \frac{1}{2} : 1 : \frac{1}{3}$.
Multiplying by $6$,we get $3 : 6 : 2$.

(D) Let the constant rate of heat supply be $\frac{dQ}{dt} = K$.
From the principle of calorimetry,$dQ = ms \Delta \theta$.
Since $\theta_0 = 0$,we have $dQ = ms \theta$.
Integrating or considering the rate,$K t = ms \theta$,which gives $\theta = \left( \frac{K}{ms} \right) t$.
Comparing this with the equation of a line $y = mx$,the slope is $\text{slope} = \frac{K}{ms} = \tan \alpha$,where $\alpha$ is the angle the graph makes with the time axis.
Thus,$S = \frac{K}{m \tan \alpha}$.
For body $A$ (mass $m$,angle $60^{\circ}$): $S_A = \frac{K}{m \tan 60^{\circ}} = \frac{K}{m \sqrt{3}}$.
For body $B$ (mass $m$,angle $45^{\circ}$): $S_B = \frac{K}{m \tan 45^{\circ}} = \frac{K}{m}$.
For body $C$ (mass $\sqrt{3}m$,angle $30^{\circ}$): $S_C = \frac{K}{(\sqrt{3}m) \tan 30^{\circ}} = \frac{K}{\sqrt{3}m \times \frac{1}{\sqrt{3}}} = \frac{K}{m}$.
Comparing the values,we get $S_B = S_C = \frac{K}{m}$ and $S_A = \frac{K}{m \sqrt{3}}$.
Since $\sqrt{3} > 1$,it follows that $\frac{K}{m \sqrt{3}} < \frac{K}{m}$.
Therefore,$S_A < S_B = S_C$,which is equivalent to $S_B = S_C > S_A$.

(A) The heat supplied to the copper is given by $Q = m_C c_C \Delta \theta_C$.
Given: $m_C = 50\,g = 0.05\,kg$,$c_C = 420\,J\,kg^{-1}\,^{\circ}C^{-1}$,$\Delta \theta_C = 10\,^{\circ}C$.
$Q = 0.05 \times 420 \times 10 = 210\,J$.
Now,the same quantity of heat $Q$ is given to $10\,g$ of water $(m_W = 0.01\,kg)$.
Using $Q = m_W c_W \Delta \theta_W$,where $c_W = 4200\,J\,kg^{-1}\,^{\circ}C^{-1}$:
$\Delta \theta_W = \frac{Q}{m_W c_W} = \frac{210}{0.01 \times 4200} = \frac{210}{42} = 5\,^{\circ}C$.

(D) Water is used as a coolant in car radiators because it has a very high specific heat capacity $(4200 \, J/kg \cdot K)$.
This property allows water to absorb a large amount of heat energy from the engine with a relatively small rise in its own temperature,making it highly efficient at transferring heat away from the engine block.

(C) Thermal capacity $(C')$ is defined as the amount of heat required to raise the temperature of a body by $1^{\circ}C$ or $1 \ K$.
It is mathematically expressed as $C' = m \times s$,where $m$ is the mass of the body and $s$ is the specific heat capacity of the material.
Since the specific heat capacity $(s)$ is a property of the material,the thermal capacity of a specific body depends directly on its mass $(m)$.

(D) Power of the drilling machine,$P = 10\, kW = 10 \times 10^{3}\, W$.
Mass of the aluminium block,$m = 8.0\, kg = 8 \times 10^{3}\, g$.
Time for which the machine is used,$t = 2.5\, min = 2.5 \times 60 = 150\, s$.
Specific heat of aluminium,$c = 0.91\, J\, g^{-1}\, K^{-1}$.
Total energy supplied by the drilling machine $= P \times t = 10 \times 10^{3} \times 150 = 1.5 \times 10^{6}\, J$.
Since $50\, \%$ of the power is lost,the useful energy $\Delta Q$ used to heat the block is $50\, \%$ of the total energy.
$\Delta Q = 0.50 \times 1.5 \times 10^{6} = 7.5 \times 10^{5}\, J$.
Using the formula $\Delta Q = m c \Delta T$,where $\Delta T$ is the rise in temperature:
$\Delta T = \frac{\Delta Q}{m c} = \frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91} = \frac{750}{8 \times 0.91} = \frac{750}{7.28} \approx 103\, ^{\circ}C$.
Thus,the rise in temperature of the block is $103\, ^{\circ}C$.

(D) The formula for heat $(Q)$ given to or taken from a substance is $Q = mc\Delta T$.
Here, $m$ is the mass of the substance, $c$ is the specific heat capacity of the substance, and $\Delta T$ is the change in temperature.
Thus, heat depends on these three factors.
Therefore, the correct option is $D$.

(A) The energy produced by burning coal is determined by its calorific value.
The calorific value of coal is approximately $30 \, MJ/kg$ or $3 \times 10^7 \, J/kg$.
Therefore,burning $1\, kg$ of coal produces $3 \times 10^7 \, J$ of energy.
Thus,the correct option is $A$.

(A) The quantity of heat required to warm a given substance depends on its mass $(m)$,the change in temperature $(\Delta T)$,and the nature of the substance.
The change in temperature of a substance,when a given quantity of heat is absorbed or rejected by it,is characterized by a quantity called the heat capacity of that substance.
Heat capacity $(S)$: Heat capacity is defined as the ratio of the heat supplied to a substance $(\Delta Q)$ to the corresponding change in its temperature $(\Delta T)$.
Mathematically,it is expressed as:
$S = \frac{\Delta Q}{\Delta T}$
Where $\Delta Q$ is the amount of heat supplied to the substance to change its temperature from $T$ to $T + \Delta T$.
The value of heat capacity depends on the type of material and its mass. Heat capacity of substances of the same material but different mass can be different.
The $SI$ unit of heat capacity is $J K^{-1}$ or $J/K$.

(D) The quantity of heat $Q$ required to change the temperature of a substance is given by the formula $Q = mc\Delta T$,where $m$ is the mass of the substance,$c$ is the specific heat capacity (which depends on the nature of the substance),and $\Delta T$ is the change in temperature.
$1$. Mass $(m)$: The heat required is directly proportional to the mass of the substance. Doubling the mass requires double the heat for the same temperature rise.
$2$. Change in temperature $(\Delta T)$: The heat required is directly proportional to the change in temperature. Doubling the temperature rise requires double the heat for the same mass.
$3$. Nature of the substance $(c)$: Different substances have different specific heat capacities. For the same mass and same temperature rise,different materials require different amounts of heat.
Therefore,the required amount of heat depends on all three factors: mass,change in temperature,and the nature of the substance.

(N/A) Specific heat capacity is defined as the amount of heat energy required to change the temperature of a unit mass of a substance by one unit (degree).
If $\Delta Q$ is the amount of heat absorbed or rejected by a substance of mass $m$ undergoing a temperature change $\Delta T$,then the specific heat capacity $s$ is given by:
$s = \frac{\Delta Q}{m \Delta T}$
Rearranging this,we get the equation: $\Delta Q = m s \Delta T$.
Specific heat capacity is a characteristic property of a substance. It depends on the nature of the substance and its temperature.
The $SI$ unit of specific heat capacity is $J \ kg^{-1} \ K^{-1}$.
The dimensional formula is $[M^0 L^2 T^{-2} K^{-1}]$.

(D) The heat $Q$ required to increase the temperature of a substance depends on the following factors:
$1$. Mass of the substance $(m)$: $Q \propto m$
$2$. Change in temperature $(\Delta T)$: $Q \propto \Delta T$
$3$. Nature of the substance,represented by its specific heat capacity $(c)$: $Q \propto c$
Combining these factors,we get the formula: $Q = mc\Delta T$.
Therefore,the heat required depends on all the given factors.

(N/A) Heat capacity $(C)$ is defined as the amount of heat energy required to raise the temperature of a given body by $1 \ K$ (or $1 \ ^\circ C$).
Mathematically, $C = \frac{Q}{\Delta T}$, where $Q$ is the heat energy and $\Delta T$ is the change in temperature.
$1$. $SI$ unit: Since heat energy $(Q)$ is measured in Joules $(J)$ and temperature $(\Delta T)$ in Kelvin $(K)$, the $SI$ unit of heat capacity is $J/K$ or $J \cdot K^{-1}$.
$2$. Dimensional formula: The dimension of heat energy $(Q)$ is $[ML^2T^{-2}]$ and the dimension of temperature $(\Delta T)$ is $[K]$.
Therefore, the dimensional formula of heat capacity is $\frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$.

(N/A) Molar specific heat is defined as the amount of heat energy required to raise the temperature of $1 \ mole$ of a substance by $1 \ K$ (or $1 \ ^\circ C$).
Mathematically,it is expressed as $C = \frac{1}{n} \frac{dQ}{dT}$,where $n$ is the number of moles,$dQ$ is the heat supplied,and $dT$ is the change in temperature.
The $SI$ unit of molar specific heat is $\text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.

(A) Water is used as a coolant in automobile radiators primarily because it has a very high specific heat capacity $(c = 4186 \ J/kg \cdot K)$.
This means that water can absorb a large amount of heat energy for a relatively small increase in its temperature.
Because of this property,water is highly effective at carrying heat away from the engine block to the radiator,where it can be dissipated into the atmosphere.
Additionally,water is readily available,inexpensive,and has a high thermal conductivity,making it an ideal heat transfer fluid.

(N/A) The specific heat capacity of water is significantly higher than that of land. During summer,water absorbs heat and its temperature increases much more slowly than land. Consequently,the air above the sea remains at a lower temperature compared to the air above the land. Therefore,when this air moves from the sea towards the land,it is felt as a cool breeze.

(B) The specific heat capacity of water is approximately $4186 \ J/(kg \cdot K)$,whereas the specific heat capacity of sand is approximately $830 \ J/(kg \cdot K)$.
Since $4186 \ J/(kg \cdot K) > 830 \ J/(kg \cdot K)$,water has a significantly higher specific heat capacity than sand.