91Ó°ÊÓ

An ice bag contains $275g$of ice at $0.0°C$used to treat sore muscles.

The liquid water had a temperature of $24.0°C$ when the bag was removed and melted.

To calculate the amount of kilojoules of heat absorbed:$24.0°C$

The heat needed to melt ice to water is calculated as:

$Heat=mass×Heatoffusion$

The factor for joules to kilojoules is

$1000J=1kJ$

As,

$\frac{1\mathrm{kJ}}{1000\mathrm{J}}:$$\frac{1000\mathrm{J}}{1\mathrm{kJ}}$

Substituting as

$Heat=275\mathrm{g}\mathrm{ice}×\frac{334J}{1\mathrm{g}\text{ice}}×\frac{1\mathrm{kJ}}{1000J}$

$Heat=91.9kJ$

The temperature change as,

$\mathrm{Δ}T=T_{\text{final}}-T_{\text{initial}}$

$$\begin{gathered} ∆T=24.0°C-0°C \\ ∆T=24.0°C \end{gathered}$$

To heat equation as:

$Heat=mass×\mathrm{Δ}T×SH$

$$\begin{gathered} Heat=275g×24.0°\mathrm{C}×4.184\frac{J}{g{}^{∘}\mathrm{C}}×\frac{1\mathrm{kJ}}{1000J} \\ Heat=27.6kJ. \end{gathered}$$

As a result ,from a temperature of $0°C$to $24°C$,the heat gained by the water is $27.6kJ.$

The amount of energy needed as:

$$\begin{gathered} Totalenergy=91.9kJ+27.6kJ \\ Totalenergy=119.5kJ. \end{gathered}$$

So, the absorbed heat is $119.5kJ$.