Question

The specific heat capacity of a metal at low temperature (T) is given as : $C_{p} \left(kjK^{-1} kg^{-1}\right) = 32 \left(\frac{T}{400}\right)^{3}$ A 100 gram vessel of this metal is to be cooled from $20^{\circ}K$ to $4^{\circ}K$ by a special refrigerator operating at room temperature $\left(27^{\circ}C\right)$. The amount of work required to cool the vessel is :

• A. greater than 0.148 kJ
• B. between 0.148 kJ and 0.028 kJ
• C. less than 0.028 kJ
• D. equal to 0.002 kJ

Answer 1

Answer: (D)

equal to 0.002 kJ

Answer 2

$Q = \int\,mcdT$ $=\int\limits^{4}_{20} 0.1\times 32 \times\left(\frac{T^{3}}{400^{3}}\right)dt$ $\approx 0.002 \,k \,J.$