Q. 3.95 - An Introduction to General, Organic, and Biological Chemistry

Question

A hot-water bottle for a patient contains $725 g$ of water at ${65}^{\circ }C$. If the water cools to body temperature $\left({37}^{\circ }C\right)$, how many kilojoules of heat could be transferred to sore muscles? 

Step-by-Step Solution

Answer

The Heat transferred from water bottle to sore muscles in Kilojoules is $84.9 KJ$

Step1: Introduction.

Any or all of numerous types of events, termed processes, that transport energy and entropy from one point to another are referred to as heat transfer. The terminology used to describe the various methods are dispersion, thermal radiation, and conduction.

Given Information's are as follows:

Mass of water$=725 g$

Initial temperature$={65}^{\circ }C$

Cool to body temperature at ${37}^{\circ }C$

Step2: Find the Heat transferred from water bottle to sore muscles.

The heat is represented by,

$q_{\text{soln }}=C_{\mathrm{s}}\times m\times \Delta T$

where,

$q_{\text{soln }}$is heat of solution, $C_{\mathrm{s}}$ is specific heat and $\Delta T$ is change in Temperature. Then,

$C_{\mathrm{s}}=4.18 \mathrm{J}/\mathrm{gC}$

$$\begin{gathered} ∆T=65C-37C \\ =28C \end{gathered}$$

Given, $m=725 g$

Substitute the values,

$q_{\text{soln }}=C_{\mathrm{s}}\times m\times \Delta T$

$$\begin{gathered} =4.18\frac{J}{g C}\times 725 g\times 28C \\ =8.49\times {10}^4 J \end{gathered}$$

Convert this into Kilojoules,

$8.49\times {10}^4 \mathrm{J}\times \frac{1 \mathrm{kJ}}{{10}^3 \mathrm{J}}=84.9 \mathrm{kJ}$

The Heat transferred to sore muscles is $84.9 \mathrm{kJ}$