Q. 3.101 - An Introduction to General, Organic, and Biological Chemistry

Question

An ice bag containing $275 \mathrm{g}$ of ice at $0.0°\mathrm{C}$ was used to treat sore muscles. When the bag was removed, the ice had melted and the liquid water had a temperature of $24.0°\mathrm{C}$. How many kilojoules of heat were absorbed? $(3.6,3.7)$

Step-by-Step Solution

Answer

The amount of heat absorbed is $119.5 kJ$.

Step 1: Given Data :

An ice bag contains $275 g$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:$24.0°C$

Step 2: Calculating heat

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

$Heat =mass \times Heat of fusion$

The factor for joules to kilojoules is

$1000 J=1 kJ$

As,

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

Substituting as

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

$Heat= 91.9 kJ$

Step 3: Change in temperature:

The temperature change as,

$\Delta T=T_{\text{final }}-T_{\text{initial }}$

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

Step 4: Heat absorbed by water:

To heat equation as:

$Heat=mass\times \Delta T\times SH$

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

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

Step 5: Total energy required to melt ice:

The amount of energy needed as:

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

So, the absorbed heat is $119.5 kJ$.