# What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve

Dec 18, 2015

The answer is (C) $\text{6680 J}$.

#### Explanation:

In order to be able to answer this question, you need to know the value of water's enthalpy of fusion, $\Delta {H}_{f}$.

As you know, a substance's enthalpy of fusion tells you how much heat is needed to melt $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$.

Simply put, a substance's enthalpy of fusion tells you how much heat is required to get $\text{1 g}$ of water to undergo a solid $\to$ liquid phase change.

Water's enthalpy of fusion is approximately equal to

$\Delta {H}_{f} = 334 \text{J"/"g}$

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

This tells you that in order to melt $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provide it with $\text{334 J}$ of heat.

In your case, you have $\text{20.0 g}$ of ice to melt, so you will need $20$ times more heat than you would need for $\text{1 g}$ of ice.

Therefore, you can say that

20.0 color(red)(cancel(color(black)("g ice"))) * "334 J"/(1color(red)(cancel(color(black)("g ice")))) = "6680 J"

are needed to melt $\text{20.0 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$.

Therefore, (C) $\text{6680 J}$ is your answer.