Question

A sports trainer applies an ice bag to the back of an injured athlete. Calculate the beat, in kilocalories, that is absorbed if $145g$of ice at role="math" localid="1651726770488" $0.0^{\circ }C$ is placed in an ice bag, melts, and warms to body temperature of $37.0^{\circ }C$.

Short answer

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

Step-by-step solution

Given Information.

Given Information are as follow as:

Mass of ice$=145g$

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

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

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 $145g$

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

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

$$\begin{gathered} Heatchange=mass\times heatoffusion \\ =145g\times 80cal/g \\ =11600cal \end{gathered}$$

Find the Heat Absorbed. 

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

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

Calculate the Total Energy. 

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

The Total Heat absorbed will be$17.0Kcal$