Question

An ice-cube tray of negligible mass contains$\mathbf{0}\mathbf{.}\mathbf{290}\mathit{k}\mathit{g}$of water at$\mathbf{18}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$. How much heat must be removed to cool the water to$\mathbf{0}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}$and freeze it? Express your answer in joules, calories, and Btu.

Short answer

The heat released will be $-1.43\times {10}^5\mathrm{J}=-3.42\times {10}^4\mathrm{cal}=135.5Btu$.

Step-by-step solution

Stages in the process

There will be two stages in the given process, the first stage includes a decrease in the temperature of the water from $18°\mathrm{C}\mathrm{to}0°\mathrm{C}$without any change in the phase, the second stage is the change in the phase from liquid (water) to solid (ice). Hence, the amount of heat that must be removed to cool the water $0°C$to and freeze it can be

${\mathrm{Q}}_{\mathrm{removed}}={\mathrm{mc}}_{\mathrm{water}}\mathrm{\Delta T}-{\mathrm{mL}}_{\mathrm{f}}$

Here, the negative sign implies the removal of heat energy.

Use the equation to find the heat removed

$\begin{array}{r}{Q}_{\text{Temoned}}=[0.35\times 4190\times (0-18\left)\right]-\left(0.35\times 33.4\times {10}^4\right)\\ Q_{\text{renoved}}=-1.43\times {10}^5\mathrm{J}\end{array}$

The values in calories and Btu will be

$\begin{array}{r}{Q}_{\text{renюегed}}=-1.43\times {10}^5\mathrm{J}\times \frac{1\mathrm{cal}}{4.186\mathrm{J}}=3.42\times {10}^4\mathrm{cal}\\ Q_{\text{remored}}=-1.43\times {10}^{5}J\times \frac{1Btu}{1055\mathrm{J}}=-135.5\mathrm{Btu}\end{array}$