# Question 27012

May 3, 2017

$1.50 \cdot {10}^{4} \text{kJ}$

#### Explanation:

You need to know the value of water's enthalpy of fusion, $\Delta {H}_{\text{fus}}$

$\Delta {H}_{\text{fus" = "333.55 J g}}^{- 1}$

https://en.wikipedia.org/wiki/Enthalpy_of_fusion

The enthalpy of fusion tells you the amount of heat needed to convert $\text{1 g}$ of a substance from solid at its melting point to liquid at its melting point.

For water, you need $\text{333.55 J}$ to convert $\text{1 g}$ of water from solid at ${0}^{\circ} \text{C}$, i.e. from ice, to liquid at ${0}^{\circ} \text{C}$.

This implies that $\text{45.0 g}$ of ice will require

45.0 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "15009.75 J"#

Expressed in kilojoules and rounded to three sig figs, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat needed" = 1.50 * 10^4color(white)(.)"kJ}}}}$