Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

AIEEEAIEEE 2011Thermal Properties of Matter

Solution:

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