Question

The specific heat capacity of a metal at low temperature $\left(T\right)$ is given as $C_P\left({kJk}^{-1}\right)=32{\left(\frac{T}{400}\right)}^2$. A $100$ gram vessel of this metal is to be cooled from ${20}^{o}K$ to $4^{o}K$ by a special refrigerator operating at room temperature $\left({27}^{o}C\right)$. 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

The correct option is A equal to $0.002kJ$

It is given that $C_p=32{\left(\frac{T}{400}\right)}^3$

Since, it is an isothermal process because here temperature is constant i.e. ${27}^{0}C$.

$Q=U+W$

$\because U=0$

$so,Q=W.......\left(1\right)$

$anddQ=m{C}_{p}dT$

$so,W=Q=\int m{C}_{p}dT$

$=\underset{20}{\overset{4}{\int }}0.1\times 32\times {\left(\frac{T}{400}\right)}^{3}dT$

$=\underset{20}{\overset{4}{\int }}\frac{3.2}{64\times {10}^6}{T}^{3}dT$

$=5\times {10}^{-8}\underset{20}{\overset{4}{\int }}{T}^{3}dT$

$=5\times {10}^{-8}\underset{20}{\overset{4}{\int }}\frac{T^4}{4}dT$

$=0.002kJ$