Q. 3.56 - An Introduction to General, Organic, and Biological Chemistry

Question

A sports trainer applies an ice bag to the back of an injured athlete. Calculate the beat, in kilocalories, that is absorbed if $145 g$ of ice at $0.0^{\circ }C$ is placed in an ice bag, melts, and warms to body temperature of $37.0^{\circ }C$.

Step-by-Step Solution

Answer

The total beat in Kilocalories if it absorbed $145 g$ of ice and warms to body temperature of $37.0^{\circ }C$ is $17 Kcal$

Step1: Given Information.

Given Information are as follow as:

Mass of ice$=145 g$

Initial temperature$=0.0^{\circ }C$

Warms to temperature$=37.0^{\circ }C$

Step2: Find the Heat Change.

The total energy is equal to the sum of the heat required to melt the ice and the heat required to raise the temperature to ${37}^{\circ }C$

Mass of ice is $145 g$

Heat of fusion of water is $80 cal/g$

Calculate the energy required to melt $145 g$ of ice using the following formula:

$$\begin{gathered} Heat change=mass\times heat of fusion \\ =145 g\times 80 cal/g \\ =11600 cal \end{gathered}$$

Step3: Find the Heat Absorbed.

Calculate the energy required to warm the liquid from $0^{\circ }C$ to ${37}^{\circ }C$ using the following formula:

$$\begin{gathered} Heat=mass\times temperature change\times specific heat \\ =145 g\times \left(37.0^{\circ }C-0^{\circ }C\right)\times 1 cal/g^{\circ }C \\ =5365 cal \end{gathered}$$

Step4: Calculate the Total Energy.

$$\begin{gathered} Total Heat energy=11600 cal+5365 cal \\ =16965 cal\times \frac{1 Kcal}{1000 cal} \\ =17.0 Kcal \end{gathered}$$

The Total Heat absorbed will be $17.0 Kcal$