## Materials Science and Engineering: An Introduction

a. $c_{p (copper)} = 24.53 \frac{J}{mol.K}$ b. $c_{p (iron)} = 25.02 \frac{J}{mol.K}$ c. $c_{p (gold)} = 25.21 \frac{J}{mol.K}$ d. $c_{p (nickel)} = 26.00 \frac{J}{mol.K}$
Required: Room temperature heat capacities $c_{p}$ of a. Copper b. Iron c. Gold d. Nickel Solution: From Table 19.1, $c_{p}$ values of the given materials can be found, and the atomic weights are found on the front cover. Room temperature heat capacities can be calculated by multiplying the $c_{p}$ with the atomic weights. a. $c_{p (copper)} = (386 \frac{J}{kg.K}) (63.55 \frac{g}{mol}) (\frac{1 kg }{1000 g}) = 24.53 \frac{J}{mol.K}$ b. $c_{p (iron)} = (448 \frac{J}{kg.K}) (55.85 \frac{g}{mol}) (\frac{1 kg }{1000 g}) = 25.02 \frac{J}{mol.K}$ c. $c_{p (gold)} = (128 \frac{J}{kg.K}) (196.97 \frac{g}{mol}) (\frac{1 kg }{1000 g}) = 25.21 \frac{J}{mol.K}$ d. $c_{p (nickel)} = (443 \frac{J}{kg.K}) (58.69 \frac{g}{mol}) (\frac{1 kg }{1000 g}) = 26.00 \frac{J}{mol.K}$