# Question 319f9

Jun 16, 2017

$\text{600 g}$

#### Explanation:

Start by looking up the enthalpy of fusion of water

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

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

Now, the enthalpy of fusion of a given substance tells you the enthalpy change that occurs when $\text{1 g}$ of said substance goes from solid to liquid at its melting point.

In your case, you know that in order to convert $\text{1 g}$ of solid water, i.e. ice, from solid at its normal melting point of ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provide it with $\text{333.55 J}$ of heat.

Now, you know that you are providing an unknown mass of ice with

200 color(red)(cancel(color(black)("kJ"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) = 2 * 10^5 $\text{J}$

of heat. In order to determine the mass of ice at ${0}^{\circ} \text{C}$ that can be converted to liquid water at ${0}^{\circ} \text{C}$, i.e. you can use the enthalpy of fusion as a conversion factor.

2 * 10^5 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(333.55color(red)(cancel(color(black)("J")))))^(color(blue)("equivalent to" color(white)(.)DeltaH_"fus" = "333.55 J g"^(-1))) = "599.6 g"#

Rounded to one significant figure, the number of sig figs you have for the amount of heat you provide to the sample, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of ice = 600 g}}}}$

You can thus say that $\text{200 kJ}$ of heat are enough to convert $\text{600 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$.