## Materials Science and Engineering: An Introduction

$c_{v}$ at 40 K= $47.84 \frac{J}{kg.K}$
Given: Copper with heat capacity ($C_{v}) = 0.38 \frac{g}{mol.K}$ at temperature of 20 K Debye Temperature = 340 K Required: specific heat at 40 K Solution: Since the temperature of the given $c_{v}$ (20 K) is significantly below the Debye temperature (340 K), Equation 19.2 may be used. $C_{v} = AT^{3}$ $A = \frac{C_{v}}{T^{3}} = \frac{0.38 \frac{J}{mol.K}}{(20 K)^{3}} = 4.75 \times 10^{-5} \frac{J}{mol.K^{4}}$ Thus at 40 K, $C_{v} = AT^{3} = (4.75 \times 10^{-5} \frac{J}{mol.K^{4}}) (40 K)^{3} = 3.04 \frac{J}{mol.K}$ $c_{v} = (3.04 \frac{J}{mol.K}) (\frac{1 mol}{63.55 g} (\frac{1000 g}{1 kg}) = 47.84 \frac{J}{kg.K}$