## Materials Science and Engineering: An Introduction

$c_{v} = 391.82 \frac{J}{kg.K}$
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 400 K Solution: Since the required temperature of 400 K is above the given Debye temperature of 340 K, use the following equation to estimate $c_{v}$, $C_{v} = 3R = (3) (8.31 \frac{J}{mol.K}) = 24.9 \frac{J}{mol.K}$ Thus at 400 K: $c_{v} = (24.9 \frac{J}{mol.K}) (\frac{1 mol}{63.55 g} (\frac{1000 g}{1 kg}) = 391.82 \frac{J}{kg.K}$