# How much heat does a 23.0 gram ice cube absorb as its temperature increases from-17.4 °C to 0.0 °C ? The specific heat of ice is 2.108 J/(g C).

Dec 24, 2016

Here's what I got.

#### Explanation:

The specific heat of ice tells you how much heat must be absorbed by $\text{1 g}$ of ice in order for its temperature to increase by ${1}^{\circ} \text{C}$.

You know that since

$c = {\text{2.108 J g"^(-1)"C}}^{- 1}$

you need to provide $\text{2.108 J}$ of heat per gram of ice in order to increase the temperature of the sample by ${1}^{\circ} \text{C}$. This means that in order to increase the temperature of $\text{23.0 g}$ of ice by ${1}^{\circ} \text{C}$, you need to provide

23.0 color(red)(cancel(color(black)("g"))) * "2.108 J"/(1color(red)(cancel(color(black)("g"))) 1^@"C") = "48.484 J g"^(-1)

So, you know that for a $\text{23.0-g}$ sample of ice, $\text{48.484 J}$ are needed in order to increase its temperature by ${1}^{\circ} \text{C}$. In your case, you need the temperature of the sample to increase by

$\Delta T = {0.0}^{\circ} \text{C" - (-17.4^@"C") = 17.4^@"C}$

This means that you will need

$17.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * "48.484 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("844 J}}}}$

I'll leave the answer rounded to three sig figs.

Keep in mind that this much heat is needed in order to get you from ice at $- {17.4}^{\circ} \text{C}$ to ice at ${0.0}^{\circ} \text{C}$. If you want the melt the ice to liquid at ${0.0}^{\circ} \text{C}$, you're going to need to provide it with additional heat.