# Question #9db85

May 20, 2017

$9 \cdot {10}^{5} \text{g}$

#### Explanation:

In order to be able to answer this question, you must know the enthalpy of fusion of water

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

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

In your case, you know that in order to convert $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provide $\text{333.55 J}$.

You can thus say that $3 \cdot {10}^{8}$ $\text{J}$ of heat will melt

$3 \cdot {10}^{8} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) * "1 g ice"/(333.55color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(9 * 10^5color(white)(.)"g ice}}}}$

The answer is rounded to one significant figure.

If you want, you can convert this to kilograms by using the fact that