# It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water?

Mar 3, 2017

$\text{240 cal}$

#### Explanation:

The idea here is that the problem is providing you with a measure of how much heat is needed per gram of solid water, i.e. of ice, in order for a solid $\to$ liquid phase change to take place.

Simply put, you are given a measure of how much heat is required to melt $\text{1 g}$ of ice at its normal melting point of ${0}^{\circ} \text{C}$.

So, you can say that ${\text{80 cal g}}^{- 1}$ is equivalent to saying that it takes $\text{80 cal}$ of heat for every $\text{1 g}$ of ice to convert it from solid at ${0}^{\circ} \text{C}$ to lqiuid water at ${0}^{\circ} \text{C}$.

This means that $\text{3 g}$ of ice would require

$3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal}}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the mass of ice.

So remember this notation

$\textcolor{b l u e}{{\text{80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g}}^{- 1}}$

means that every $\textcolor{red}{\text{1 g}}$ of ice at ${0}^{\circ} \text{C}$ can be converted to liquid water at ${0}^{\circ} \text{C}$ by providing it with $\textcolor{b l u e}{\text{80 cal}}$ of heat.