Animated Solution for Physics - Thermodynamics: The specific heat capacity of a metal at low temperature ($T$) is given as $C_p({\text{kJ K}}^{-1}{\text{kg}}^{-1})=32{\left(\frac{T}{400}\right)}^3$. A 100 g vessel of this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is

This problem is a beautiful intersection of calculus and thermodynamics. It challenges the common assumption that specific heat and sink temperatures are always constant. Let's break down the physics step-by-step.

Usually, when we calculate the heat required to change an object's temperature, we use the familiar formula $Q=m{C}_p\Delta T$. However, this formula only works if the specific heat capacity $C_p$ is a constant.

In this problem, we are operating at extremely low temperatures (near absolute zero). At these temperatures, the specific heat of metals drops drastically and is proportional to the cube of the temperature, given by $C_p=32{\left(\frac{T}{400}\right)}^3$. Because $C_p$ is continuously changing, we must use integration to sum up the infinitesimally small amounts of heat $dQ$ extracted at each temperature $T$.

Substituting the given values ($m=0.1$ kg, and integrating from 4 K to 20 K to find the total positive heat extracted):

Evaluating this integral yields the total heat extracted from the vessel:

Now, we need to find the work $W$ done by the refrigerator to extract this heat. For an ideal Carnot refrigerator, the Coefficient of Performance (COP) is defined as the ratio of heat extracted from the cold sink ($Q_2$) to the work input ($W$):

Rearranging for work, we get:

Here is the critical catch: The sink temperature $T_2$ is not constant! It is dropping from 20 K down to 4 K. As $T_2$ decreases, the temperature gap between the room ($T_1=300$ K) and the vessel widens. Thermodynamics dictates that pumping heat across a larger temperature gap requires more work.

Instead of performing a complex integral for the work, we can use a clever bounding technique. We calculate the work required if all the heat was extracted at the easiest temperature (20 K), and the work required if all the heat was extracted at the hardest temperature (4 K).

Minimum Work Bound (at $T_2=20$ K):

Maximum Work Bound (at $T_2=4$ K):

Since the heat is actually extracted continuously as the temperature drops from 20 K to 4 K, the true total work done must lie strictly between these two extremes.

Therefore, the required work is between 0.148 kJ and 0.028 kJ.