Question #d5e47

Jan 11, 2016

$3030 k J$

Explanation:

Your method is 100% correct, but you made a small calculation error in converting the lb to g, by a factor of 10, resulting in you final answer being out by a factor of 10.

The amount of energy required to melt $1 k g$ of ice at ${0}^{\circ} C$ is equal to the specific latent heat of fusion for ice and has value ${L}_{f} = 334 k J / k g$.

So for $20 \times 0.4536 = 9.072 k g$, we get that energy required to melt it is
$W = m {L}_{f} = 9.072 k g \times 334000 J / k g$

$= 3.03 M J$

$= 3030 k J$