Question

The specific heat capacity of a metal at low temperature $(T)$ is given as ${C_p}(KJ{K^{ - 1}}) = 32{\left( {\dfrac{T}{{400}}} \right)^2}$. A $100gm$ vessel of this metals to be cooled from ${20^o}K$ to ${4^o}K$ by a special refrigerator operating at room temperature$({27^o}C)$. The amount of work required to cool the vessel is:
A) Equal to $0.002KJ$
B) Greater than $0.148KJ$
C ) Between $0.148KJ$ and $0.028KJ$
D) Less than $0.028KJ$

Answer 1

This is an isothermal process, so we consider $U = 0$. Now we use the first law of thermodynamics to calculate the amount of work done.
We substitute the value of $Q$ after integrating $dQ = m{C_P}dt$.
The integration is performed within the intervals over which the cooling takes place.

Complete step-by-step answer:
The given process is an isothermal process, therefore, the temperature remains constant, and hence we consider the room temperature which is given as $({27^o}C)$.
Since the temperature is constant, we can consider change in internal energy to be zero.
Therefore, $U = 0$.
Using First law of thermodynamics, as we know:
$\Delta U = Q - W$
Therefore, we obtain:
$Q = W$
where,
$Q =$Heat added
$W =$Work done by the system.
We need to find the value of $W$ in the question.
We know:
$dQ = m{C_P}dt$
Where:
$m$ is the mass of substance
${C_p}$ is specific heat capacity at constant pressure
$dt$ is a change in temperature.
Therefore, we integrate $dQ$ to get $Q$ which is equal to $W$.
$W = Q = \smallint m{C_P}dt$
Putting the value of ${C_P}$ as given in the question, we obtain:
$W = Q = \smallint m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT$
It is given that the body cools from ${20^o}K$to ${4^o}K$, so we integrate within this interval.
Thus:
$W = Q = \smallint _{20}^4m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT$
Mass of the vessel is given as $100gm$, thus we take $m = 0.1Kg$
Thus,
$W = Q = \smallint _{20}^40.1 \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT$

Thus after solving the equation, we obtain:
$W = Q = 0.002KJ$.

Hence, option (A) is correct.

Note: Isothermal process is referred to as the process of thermodynamics which takes place at a constant temperature and thus exchange of energy takes place very slowly. This is the reason as a result of which thermal equilibrium is maintained.
Therefore, there is no heat transferred between the system and the surrounding.