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

$(12 \, erg (^{\circ} C)^{-1})$ The heat capacity $C$ is defined as the product of the mass $m$ and the specific heat capacity $s$ of the substance.
Given:
Mass $m = 40 \, g$
Specific heat $s = 0.3 \, erg \, g^{-1} (^{\circ} C)^{-1}$
Calculation:
$C = m \times s$
$C = 40 \, g \times 0.3 \, erg \, g^{-1} (^{\circ} C)^{-1}$
$C = 12 \, erg (^{\circ} C)^{-1}$

(N/A) The ratio of heat supplied $\Delta Q$ to a substance to the change in its temperature $\Delta T$ is called heat capacity.
$\therefore$ Heat capacity $= \frac{\text{Heat given}}{\text{Change in temperature}}$
$S = \frac{\Delta Q}{\Delta T}$
The $SI$ unit of heat capacity is $J K^{-1}$ or $cal K^{-1}$.
The heat capacity of a body depends on the material of the body as well as on its mass.
For different bodies,having different types of masses but of the same material,the value of heat capacity is different.

(N/A) Specific heat is defined as the heat capacity per unit mass of a body.
$s = \frac{\text{Heat capacity}}{\text{Mass of body}}$
$s = \frac{\Delta Q}{m \Delta T}$
$\therefore \Delta Q = m s \Delta T$
The $SI$ unit of specific heat is $J \cdot kg^{-1} \cdot K^{-1}$.
The magnitude of specific heat depends on the nature of the material of the body,the temperature range,and the conditions under which heat is supplied (e.g.,constant pressure or constant volume).
Heat capacity per mole of a substance is known as molar specific heat capacity $(C)$:
$C = \frac{S}{\mu} = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$
Molar specific heat capacity is independent of the amount of substance. Its unit is $J \cdot mol^{-1} \cdot K^{-1}$.

Substance	Specific Heat $(J \cdot kg^{-1} \cdot K^{-1})$	Molar Specific Heat $(J \cdot mol^{-1} \cdot K^{-1})$
Aluminium	$900.0$	$24.4$
Carbon	$506.5$	$6.1$
Copper	$386.4$	$24.5$
Lead	$127.7$	$26.5$
Silver	$236.1$	$25.5$
Tungsten	$134.4$	$24.9$

Experimental values agree with the predicted value of $3R$ at ordinary temperatures (except for Carbon). This agreement breaks down at low temperatures.

(N/A) The old unit of heat was the calorie, and one calorie was earlier defined as the amount of heat required to raise the temperature of $1 \; g$ of water by $1^{\circ} C$.
Variation of specific heat capacity of water with temperature:
The specific heat of water varies slightly with temperature; hence, for a precise definition of a calorie, it was necessary to specify the unit temperature interval.
Precise definition of a Calorie: The amount of heat required to raise the temperature of $1 \; g$ of water from $14.5^{\circ} C$ to $15.5^{\circ} C$.
The specific heat capacity of water is approximately $4186 \; J \; kg^{-1} \; K^{-1}$, which means $4.186 \; J \; g^{-1} \; K^{-1}$.
From the relation $W = JH$, the amount of work needed to produce $1 \; \text{cal}$ of heat is called the mechanical equivalent of heat.
Therefore, $W = J$ (where $H = 1 \; \text{calorie}$).
Thus, there are two units of heat, Joule and Calorie, and for conversion, $1 \; \text{calorie} = 4.186 \; J$ of heat is needed.

(N/A) Heat capacity is defined as the amount of heat energy required to raise the temperature of a given substance by $1 \ K$ (or $1 \ ^\circ C$).
Mathematically,it is given by $C = \frac{dQ}{dT}$,where $dQ$ is the heat supplied and $dT$ is the change in temperature.
The $SI$ unit of heat capacity is $\text{joule per kelvin}$ ($J/K$ or $J \cdot K^{-1}$).

(D) The heat capacity $(C)$ of a substance is defined as the amount of heat required to raise the temperature of a given mass of the substance by $1 \ K$ or $1^{\circ}C$.
It depends on several factors:
$1$. Nature of the substance: Different materials have different atomic or molecular structures,leading to different heat capacities.
$2$. Temperature of the substance: For many substances,heat capacity varies with temperature,especially at very low temperatures.
$3$. State of the substance: The heat capacity of a substance in its solid,liquid,or gaseous state is different because the intermolecular forces and degrees of freedom change with the state.
Therefore,the value of heat capacity depends on all the mentioned factors.

(N/A) Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of a unit mass of the substance by $1 \ K$ (or $1 \ ^\circ C$).
Mathematically,it is given by $s = \frac{Q}{m \Delta T}$,where $Q$ is the heat energy,$m$ is the mass,and $\Delta T$ is the change in temperature.
The $SI$ unit of specific heat capacity is $\text{Joule per kilogram per Kelvin}$ $(J \ kg^{-1} \ K^{-1})$.

(D) The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of a unit mass of the substance by $1 \ K$ or $1 \ ^\circ C$.
It depends on the following factors:
$1$. Nature of the substance: Different materials have different atomic or molecular structures,which affect how they store thermal energy.
$2$. Temperature of the substance: For many materials,specific heat capacity varies with temperature,especially at very low or very high temperatures.
$3$. State of the substance: The specific heat capacity of a substance changes when it changes its state (e.g.,solid to liquid or liquid to gas) because the internal energy and molecular arrangements are different in each state.
Therefore,the correct option is $D$.

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

(A) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2}RT$ to the internal energy of one mole of a substance.
$A$ solid atom can be modeled as a three-dimensional harmonic oscillator.
For a three-dimensional harmonic oscillator,there are $3$ degrees of freedom for kinetic energy and $3$ degrees of freedom for potential energy,totaling $6$ degrees of freedom.
Therefore,the total internal energy $U$ for one mole of a solid is $U = 6 \times (\frac{1}{2}RT) = 3RT$.

(N/A) The old definition of a calorie is the amount of heat energy required to raise the temperature of $1 \ g$ of water from $14.5 \ ^\circ C$ to $15.5 \ ^\circ C$ at a pressure of $1 \ atm$.
The new definition of a calorie is defined in terms of the Joule,which is the $SI$ unit of energy. One calorie is exactly equal to $4.184 \ J$.

(N/A) The specific heat capacity of water $(c)$ is not constant; it varies with temperature.
At $0 \ ^\circ C$,the specific heat capacity of water is approximately $4.217 \ J/g \cdot K$.
As the temperature increases,the specific heat capacity decreases,reaching a minimum value of approximately $4.178 \ J/g \cdot K$ at around $35 \ ^\circ C$.
Beyond $35 \ ^\circ C$,the specific heat capacity starts to increase again as the temperature rises toward $100 \ ^\circ C$.
The graph of specific heat capacity versus temperature is a $U$-shaped curve,showing a minimum at room temperature.

(N/A) The specific heat capacity of water is the amount of heat required to raise the temperature of $1 \ kg$ of water by $1 \ K$ (or $1 \ ^\circ C$).
The value of the specific heat capacity of water is approximately $4186 \ J \ kg^{-1} \ K^{-1}$ or $4.186 \ J \ g^{-1} \ ^\circ C^{-1}$.
In terms of calories,it is $1 \ cal \ g^{-1} \ ^\circ C^{-1}$.

(B) The specific heat capacity of water is not constant and varies with temperature.
Experimental data shows that the specific heat capacity of water decreases as the temperature increases from $0 \ ^\circ C$ to approximately $35 \ ^\circ C - 40 \ ^\circ C$.
It reaches a minimum value at approximately $35 \ ^\circ C$ to $37 \ ^\circ C$.
Among the given options,$35 \ ^\circ C$ is the standard value often cited in physics textbooks as the temperature where the specific heat of water is at its minimum.

(N/A) No, the specific heat capacities of water and ice are not the same.
The specific heat capacity of water is approximately $4.18 \, J \, g^{-1} \, K^{-1}$ (or $1 \, cal \, g^{-1} \, ^\circ C^{-1}$).
The specific heat capacity of ice is approximately $2.09 \, J \, g^{-1} \, K^{-1}$ (or $0.5 \, cal \, g^{-1} \, ^\circ C^{-1}$).

(N/A) In solids, atoms oscillate about their mean position.
Let the number of atoms in $1$ mole of a solid be $N_{A}$.
According to the law of equipartition of energy, the energy associated with the oscillation of atoms in one dimension is $2 \times \frac{1}{2} k_{B} T = k_{B} T$.
Therefore, in three dimensions, the average energy of an atom is $3 k_{B} T$.
The total energy of $1$ mole of solid is $U = 3 k_{B} T \times N_{A} = 3 RT$ (since $k_{B} N_{A} = R$).
From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.
Since $\Delta W = P \Delta V$ and for solids $\Delta V \approx 0$, we have $\Delta Q = \Delta U$.
The molar specific heat of a solid is $C = \frac{\Delta Q}{\Delta T} = \frac{d(3 RT)}{dT} = 3R$.
Substituting $R = 8.31 \ J \ mol^{-1} \ K^{-1}$, we get $C = 3 \times 8.31 = 24.93 \ J \ mol^{-1} \ K^{-1}$.

Substance	Specific Heat $(J \ kg^{-1} \ K^{-1})$	Molar Specific Heat $(J \ mol^{-1} \ K^{-1})$
Aluminium	$900.0$	$24.4$
Carbon	$506.5$	$6.1$
Copper	$386.4$	$24.5$
Lead	$127.7$	$26.5$
Silver	$236.1$	$25.5$
Tungsten	$134.4$	$24.9$

(A) According to the law of equipartition of energy,each degree of freedom contributes $1/2 k_BT$ to the average energy of a particle.
In a solid,each atom acts as a 3D harmonic oscillator.
Each atom has $3$ degrees of freedom for kinetic energy and $3$ degrees of freedom for potential energy,totaling $6$ degrees of freedom.
The average energy per atom is $U = 6 \times (1/2 k_BT) = 3k_BT$.
For $1 \text{ mole}$ of a solid,the total internal energy is $U = 3RT$.
The molar specific heat at constant volume is $C_V = dU/dT = 3R$.

(A) The law of equipartition of energy is applicable to ideal gases,where the internal energy is solely due to the kinetic energy of the molecules.
Water is a liquid,and its specific heat is determined by its complex intermolecular forces and hydrogen bonding,which cannot be derived using the law of equipartition of energy.
However,the standard experimental value for the specific heat capacity of water is $1 \text{ cal/g°C}$ or $4186 \text{ J/kg°K}$.
Since the question asks for the value of the specific heat of water,the correct standard value is $1 \text{ cal/g°C}$.

(A) The specific heat capacity $C$ is defined as the amount of heat required to raise the temperature of a unit mass of a substance by $1 \ K$. Mathematically,$C = \frac{dQ}{m \ dT}$.
According to the Third Law of Thermodynamics,as the temperature $T$ approaches absolute zero $(0 \ K)$,the entropy of a perfect crystal approaches zero.
Furthermore,the heat capacity of all substances approaches zero as the temperature approaches absolute zero $(T \to 0 \ K)$.
Therefore,the specific heat of all substances at absolute zero temperature is $0$.

(A) According to the Dulong-Petit law,the molar specific heat capacity of a solid at room temperature is approximately constant and is given by $C_v = 3R$,where $R$ is the universal gas constant. Since $R \approx 2 \text{ cal/mol K}$,the value is approximately $3 \times 2 = 6 \text{ cal/mol K}$. However,the specific value $3 \text{ cal/mol K}$ mentioned in the prompt is incorrect; the correct classical value is $3R \approx 6 \text{ cal/mol K}$.

(B) The specific heat capacity of water is very high $(4186 \ J/kg \cdot K)$.
This means that water can absorb or release a large amount of heat energy with only a small change in its temperature.
Because of this property,it is highly effective at carrying heat away from engines or industrial machinery,making it an ideal coolant.

(A) Yes,the specific heat capacity of water is significantly greater than that of sand.
Water has a specific heat capacity of approximately $4186 \ J/(kg \cdot K)$,whereas sand has a specific heat capacity of approximately $800 \ J/(kg \cdot K)$.
This high specific heat capacity of water allows it to absorb or release large amounts of heat with relatively small changes in temperature,which is why it is used as a coolant and plays a crucial role in regulating Earth's climate.

(N/A) The heat capacity $(C)$ of a substance depends on two primary factors:
$1$. The nature of the substance (material properties).
$2$. The mass of the substance $(m)$.
It is defined by the relation $C = ms$,where $s$ is the specific heat capacity of the material.

(N/A) The amount of heat $Q$ required to change the temperature of a substance is given by the formula $Q = mc\Delta T$,where:
$1$. $m$ is the mass of the substance.
$2$. $c$ is the specific heat capacity of the material,which depends on the type of substance.
$3$. $\Delta T$ is the change in temperature.
Therefore,the required heat depends on the mass of the substance,the specific heat capacity (nature of the material),and the change in temperature.

(A) When a substance absorbs heat without changing its state, the relationship between the heat absorbed $(Q)$, mass $(m)$, specific heat capacity $(c)$, and the change in temperature $(\Delta T)$ is given by the formula: $Q = mc\Delta T$.
From this equation, the temperature difference $(\Delta T)$ can be determined as: $\Delta T = \frac{Q}{mc}$.
Therefore, the specific heat capacity $(c)$ is the property that relates the heat absorbed to the change in temperature when no phase change occurs.

(B) The specific heat capacity of water is very high $(4186 \ J/kg \cdot K)$ compared to most other common substances.
This means that water can absorb or release a large amount of heat energy while undergoing only a small change in temperature.
Consequently,a hot water bag remains warm for a much longer duration,providing consistent heat for therapeutic fomentation.

(N/A) Water is used as a coolant in automobile radiators primarily due to its exceptionally high specific heat capacity $(c = 4186 \ J/kg \cdot K)$.
Because of this high value,water can absorb a large amount of heat energy with a relatively small rise in its temperature.
Additionally,water is easily available,inexpensive,and has a high thermal conductivity,which allows it to efficiently transfer heat away from the engine block to the radiator fins where it is dissipated into the atmosphere.

(B) The specific heat capacity of water is much higher than that of land (soil/sand).
Due to this,water takes a longer time to heat up and a longer time to cool down compared to land.
During the day,the land heats up faster than the sea.
As a result,the air above the land becomes hot and rises,creating a low-pressure region.
Cooler air from above the sea then moves towards the land to fill this low-pressure region,which is felt as a cool sea breeze at the shores.

(B) The specific heat capacity of sand (which makes up the desert land) is very low.
Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by $1 \ ^\circ C$.
Because the specific heat of sand is low,it absorbs heat quickly during the day,leading to a rapid rise in temperature.
Similarly,it loses heat quickly at night,leading to a rapid drop in temperature.

(B) The heat required to change the temperature of a body is given by the formula $\Delta Q = M s \Delta T$,where $M$ is the mass,$s$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Since both spheres are made of copper,the specific heat capacity $s$ is the same for both.
Given that the temperature change $\Delta T = 1 \ K$ is the same for both,we have $\Delta Q \propto M$.
The mass $M$ of a sphere is given by $M = V \rho = \frac{4}{3} \pi r^3 \rho$,where $V$ is the volume,$r$ is the radius,and $\rho$ is the density.
Since $\rho$ is the same for both copper spheres,$M \propto r^3$.
Therefore,the ratio of heat required is $\frac{\Delta Q_1}{\Delta Q_2} = \frac{M_1}{M_2} = \left( \frac{r_1}{r_2} \right)^3$.
Given $r_1 = 1.5 r_2$,we have $\frac{r_1}{r_2} = 1.5 = \frac{3}{2}$.
Thus,$\frac{\Delta Q_1}{\Delta Q_2} = \left( \frac{3}{2} \right)^3 = \frac{27}{8}$.

(B) Given that the bodies are heated at a uniform rate,the rate of heat supply is constant,i.e.,$\left(\frac{\Delta Q}{\Delta t}\right)_{A} = \left(\frac{\Delta Q}{\Delta t}\right)_{B}$.
We know that $\Delta Q = mc\Delta T$,so $\frac{\Delta Q}{\Delta t} = mc\left(\frac{\Delta T}{\Delta t}\right)$.
Since the masses $m$ are equal,we have $c_{A}\left(\frac{\Delta T}{\Delta t}\right)_{A} = c_{B}\left(\frac{\Delta T}{\Delta t}\right)_{B}$.
Therefore,the ratio of specific heat capacities is $\frac{c_{A}}{c_{B}} = \frac{(\Delta T / \Delta t)_{B}}{(\Delta T / \Delta t)_{A}}$.
From the graph,the slope $(\Delta T / \Delta t)$ for body $A$ is $\frac{120 - 0}{3 - 0} = 40 \ ^{\circ}\text{C/s}$.
The slope $(\Delta T / \Delta t)$ for body $B$ is $\frac{90 - 0}{6 - 0} = 15 \ ^{\circ}\text{C/s}$.
Thus,$\frac{c_{A}}{c_{B}} = \frac{15}{40} = \frac{3}{8}$.

(B) Assuming no heat loss,the heat gained by the liquid in the calorimeter is equal to the heat supplied by the heater.
$Q = m S \Delta T = P t$
Where $m$ is the mass of the liquid,$S$ is the specific heat capacity,and $T_f - T_i = \Delta T$.
$\Rightarrow m S (T_f - T_i) = P t$
$\Rightarrow T_f = \left( \frac{P}{m S} \right) t + T_i$
Comparing this with the equation of a straight line $y = mx + c$,we get:
Slope $= \frac{P}{m S} = \frac{P}{C}$,where $C = m S$ is the heat capacity of the liquid.
Since the slope is $\frac{P}{C}$,the heat capacity $C$ is inversely proportional to the slope of the graph $(C \propto \frac{1}{\text{slope}})$.
Therefore,option $(b)$ is correct.

(B) The thermal capacity of a substance is defined as the product of its mass $(m)$ and its specific heat capacity $(c)$.
Formula: $\text{Thermal Capacity} = m \times c$
Given:
Mass $(m)$ = $100 \,g$
Specific heat $(c)$ = $0.2 \,cal / g ^{\circ} C$
Calculation:
$\text{Thermal Capacity} = 100 \,g \times 0.2 \,cal / g ^{\circ} C = 20 \,cal / ^{\circ} C$
Therefore,the correct option is $B$.

(C) The correct option is $(C)$.
Water is used as a coolant in engines primarily because it has a very high specific heat capacity $(c \approx 4186 \ J/kg \cdot K)$.
Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by $1 \ ^\circ C$.
Due to its high specific heat,water can absorb a large amount of heat energy from the engine with only a relatively small increase in its own temperature.
This makes it an extremely efficient medium for transferring heat away from engine components,preventing them from overheating.

(B) The heat energy supplied at a constant rate $H$ is given by $H = \frac{dQ}{dt} = mc \frac{dT}{dt}$,where $m$ is the mass and $c$ is the specific heat capacity.
Since $H$ is constant,we have $c = \frac{H}{m (dT/dt)}$.
This implies that the specific heat $c$ is inversely proportional to the slope of the $T-t$ graph,i.e.,$c \propto \frac{1}{\text{slope}}$.
From the figure,the slope of the graph for substance $P$ is greater than the slope of the graph for substance $Q$ (i.e.,$\text{slope}_P > \text{slope}_Q$).
Therefore,the specific heat of $Q$ must be greater than the specific heat of $P$ $(c_Q > c_P)$.
Thus,the correct statement is that the specific heat of $Q$ is greater than $P$.

(A) Given power $P = \frac{dQ}{dt}$.
Heat capacity $S = \frac{dQ}{dT} = \frac{dQ/dt}{dT/dt} = \frac{P}{dT/dt}$.
Given $T(t) = T_0(1 + \beta t^{1/4})$.
Differentiating with respect to $t$:
$\frac{dT}{dt} = T_0 \cdot \beta \cdot \frac{1}{4} t^{-3/4} = \frac{\beta T_0}{4} t^{-3/4}$.
Substituting this into the expression for $S$:
$S = \frac{P}{(\beta T_0 / 4) t^{-3/4}} = \frac{4P}{\beta T_0} t^{3/4}$.
From the given equation,$\beta t^{1/4} = \frac{T(t) - T_0}{T_0}$.
Therefore,$t^{1/4} = \frac{T(t) - T_0}{\beta T_0}$.
Raising both sides to the power of $3$:
$t^{3/4} = \left( \frac{T(t) - T_0}{\beta T_0} \right)^3$.
Substituting $t^{3/4}$ back into the expression for $S$:
$S = \frac{4P}{\beta T_0} \cdot \frac{(T(t) - T_0)^3}{\beta^3 T_0^3} = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$.

(D) The rate of heat absorption is given by $R = \frac{dq}{dt} = m C \frac{dT}{dt}$. Since the temperature is increased at a constant rate,$\frac{dT}{dt} = k$ (constant). Thus,$R \propto C$.
$(A)$ In the range $0-100 \ K$,the graph of $C$ vs $T$ is non-linear (curved),so the rate of heat absorption $R$ does not vary linearly with $T$. Statement $(A)$ is incorrect.
$(B)$ The heat absorbed is $\Delta Q = \int m C dT$,which is proportional to the area under the $C-T$ curve. The area under the curve from $0-100 \ K$ is clearly smaller than the area under the curve from $400-500 \ K$ (where $C$ is nearly constant at its maximum value). Statement $(B)$ is correct.
$(C)$ In the range $400-500 \ K$,the graph of $C$ vs $T$ is a horizontal line,meaning $C$ is constant. Since $R \propto C$,the rate of heat absorption remains constant. Statement $(C)$ is correct.
$(D)$ In the range $200-300 \ K$,the graph of $C$ vs $T$ has a positive slope,meaning $C$ is increasing. Since $R \propto C$,the rate of heat absorption increases. Statement $(D)$ is correct.
Therefore,statements $(B, C, D)$ are correct.

(D) The initial temperature is $T_i = -73^{\circ} C = (-73 + 273) \ K = 200 \ K$.
The final temperature is $T_f = 27^{\circ} C = (27 + 273) \ K = 300 \ K$.
The heat required $Q$ is given by the integral $Q = \int_{T_i}^{T_f} m C dT$.
Given $m = 1 \ kg$ and $C = kT$,we have $Q = \int_{200}^{300} (1) (kT) dT$.
$Q = k \int_{200}^{300} T dT = k \left[ \frac{T^2}{2} \right]_{200}^{300}$.
$Q = \frac{k}{2} [300^2 - 200^2] = \frac{k}{2} [90000 - 40000] = \frac{k}{2} [50000]$.
$Q = 25000 k$.
Comparing this with $nk$,we get $n = 25000$.

(A) The heat required to raise the temperature of a body is given by $Q = mc\Delta T$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Since the spheres are made of the same material (copper),$c$ is constant. For a given $\Delta T = 1 \ K$,$Q \propto m$.
The mass $m$ of a sphere is given by $m = \rho V$,where $\rho$ is the density and $V$ is the volume.
$V = \frac{4}{3}\pi r^3$,so $m \propto r^3$.
Therefore,$Q \propto r^3$.
The ratio of heat required is $\frac{Q_1}{Q_2} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$.
Given $r_1 = 1.5 r_2$,we have $\frac{r_1}{r_2} = 1.5 = \frac{3}{2}$.
Thus,$\frac{Q_1}{Q_2} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

(B) Specific heat capacity is defined as the amount of heat energy required to raise the temperature of $1 \ kg$ of a substance by $1 \ K$.
We use the formula $Q = m \cdot c \cdot \Delta T$,where $Q$ is the heat energy $(J)$,$m$ is the mass $(kg)$,$c$ is the specific heat capacity,and $\Delta T$ is the change in temperature $(K)$.
Rearranging for $c$,we get $c = \frac{Q}{m \cdot \Delta T}$.
Substituting the units,the $S.I.$ unit of specific heat capacity is $\frac{J}{kg \cdot K}$,which is written as $J \ kg^{-1} \ K^{-1}$.

(C) Given:
For copper: Mass $m_1 = 50 \text{ g} = 0.05 \text{ kg}$,Temperature rise $\Delta t_1 = 10^{\circ} \text{C}$,Specific heat $s_1 = 420 \text{ J kg}^{-1} {}^{\circ} \text{C}^{-1}$.
For water: Mass $m_2 = 10 \text{ g} = 0.01 \text{ kg}$,Specific heat $s_2 = 4200 \text{ J kg}^{-1} {}^{\circ} \text{C}^{-1}$,Let the rise in temperature be $\Delta t_2$.
Since the same amount of heat $Q$ is supplied to both,we have $Q_1 = Q_2$.
Using the formula $Q = m s \Delta t$,we get:
$m_1 s_1 \Delta t_1 = m_2 s_2 \Delta t_2$
Substituting the values:
$0.05 \times 420 \times 10 = 0.01 \times 4200 \times \Delta t_2$
$210 = 42 \times \Delta t_2$
$\Delta t_2 = \frac{210}{42} = 5^{\circ} \text{C}$.
Thus,the rise in temperature of water is $5^{\circ} \text{C}$.

(B) Given: Number of moles $n = 5$, initial temperature $T_1 = 100^{\circ} C$, final temperature $T_2 = 120^{\circ} C$, change in internal energy $\Delta U = 80 \,J$.
The change in temperature is $\Delta T = T_2 - T_1 = 120^{\circ} C - 100^{\circ} C = 20 \,K$ (since the change in temperature is the same in Celsius and Kelvin).
The change in internal energy for a gas at constant volume is given by $\Delta U = C_V \Delta T$, where $C_V$ is the total heat capacity at constant volume.
Therefore, $C_V = \frac{\Delta U}{\Delta T} = \frac{80 \,J}{20 \,K} = 4 \,J/K$.

(D) The increase in internal energy $(\Delta U)$ of a substance when heated is equal to the heat energy $(Q)$ absorbed by it, provided there is no change in state or external work done.
Using the formula for heat absorbed: $Q = m \cdot c \cdot \Delta T$
Given:
Mass $(m)$ = $5 \ kg$
Specific heat capacity $(c)$ = $4200 \ J \ kg^{-1} \ K^{-1}$
Change in temperature $(\Delta T)$ = $30^{\circ} C - 20^{\circ} C = 10^{\circ} C = 10 \ K$
Substituting the values:
$Q = 5 \ kg \times 4200 \ J \ kg^{-1} \ K^{-1} \times 10 \ K$
$Q = 210000 \ J$
$Q = 210 \ kJ$
Therefore, the increase in internal energy is $210 \ kJ$.

(B) The heat given to a substance is given by the formula $Q = mc\Delta T$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
For the metal: $Q = m_m c_m \Delta T_m = 100 \times c_m \times 20 = 2000 c_m$.
For the water: $Q = m_w c_w \Delta T_w = 20 \times c_w \times \Delta T_w$.
Since the heat given is the same,we equate the two expressions: $2000 c_m = 20 c_w \Delta T_w$.
Given the ratio of specific heat capacities $\frac{c_m}{c_w} = \frac{1}{10}$,we have $c_w = 10 c_m$.
Substituting this into the equation: $2000 c_m = 20 \times (10 c_m) \times \Delta T_w$.
$2000 c_m = 200 c_m \times \Delta T_w$.
Dividing both sides by $200 c_m$: $\Delta T_w = \frac{2000}{200} = 10^{\circ} C$.

(C) Specific heat capacity is an intrinsic property of a material,which means it depends only on the nature of the substance and not on the amount or mass of the substance.
Since the material remains copper,its specific heat capacity remains constant regardless of the change in mass.
Therefore,if the mass is doubled,the specific heat capacity will remain $s$.

(D) Given: Mass of water,$m = 2 \,kg$
Change in temperature,$\Delta T = 10^{\circ} C$
Specific heat capacity of water,$c = 4200 \,J \,kg^{-1} \,K^{-1}$
The heat energy $Q$ released is calculated using the formula:
$Q = m c \Delta T$
Substituting the values:
$Q = 2 \,kg \times 4200 \,J \,kg^{-1} \,K^{-1} \times 10 \,K$
$Q = 84000 \,J$

(D) Given: Mass of water $m = 200 \,g = 0.2 \,kg$, time $t = 200 \,s$, specific heat $s = 4200 \,J \,kg^{-1} \,K^{-1}$, and temperature change $\Delta T = 100^{\circ} C - 40^{\circ} C = 60 \,K$.
The heat energy $Q$ required to raise the temperature of the water is given by $Q = m \cdot s \cdot \Delta T$.
The power $P$ supplied by the heater is $P = \frac{Q}{t} = \frac{m \cdot s \cdot \Delta T}{t}$.
Substituting the values: $P = \frac{0.2 \,kg \times 4200 \,J \,kg^{-1} \,K^{-1} \times 60 \,K}{200 \,s}$.
$P = \frac{50400}{200} \,W = 252 \,W$.

(A) The heat absorbed by a substance is given by the formula: $H = m s \Delta T$,where $H$ is the heat,$m$ is the mass,$s$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Given values are: $H = 2100.0 \ J$,$s = 900 \ J \ kg^{-1} K^{-1}$,and $\Delta T = 2^{\circ} C = 2 \ K$.
Rearranging the formula to solve for mass $m$: $m = \frac{H}{s \Delta T}$.
Substituting the values: $m = \frac{2100}{900 \times 2} = \frac{2100}{1800} = 1.166... \ kg$.
Rounding to two decimal places,we get $m \approx 1.16 \ kg$.

(A) Given that,mass $m = 5 \,kg = 5000 \,g$.
Specific heat capacity $c = 4.2 \,J / g^{\circ} C$.
Initial temperature $T_1 = 20^{\circ} C$.
Boiling point of water $T_2 = 100^{\circ} C$.
Change in temperature $\Delta T = T_2 - T_1 = 100^{\circ} C - 20^{\circ} C = 80^{\circ} C$.
The heat energy supplied is given by the formula $\Delta Q = m c \Delta T$.
Substituting the values: $\Delta Q = 5000 \,g \times 4.2 \,J / g^{\circ} C \times 80^{\circ} C$.
$\Delta Q = 5000 \times 4.2 \times 80 = 1680000 \,J$.
Converting to kilojoules: $\Delta Q = 1680 \,kJ$.