Question

In the situation described in Exercise $\mathbf{17}\mathbf{.}\mathbf{42}$, the man’s metabolism will eventually return the temperature of his body (and of the soft drink that he consumed) to $\mathbf{37}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$. If his body releases energy at a rate of $\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{k}\mathit{J}\mathbf{/}\mathit{d}\mathit{a}\mathit{y}$(the basal metabolic rate, or BMR), how long does this take? Assume that all of the released energy goes into raising the temperature.

Short answer

The metabolism will take$7.6\min$to bring back the body to the initial temperature.

Step-by-step solution

Heat flow equation

The heat flow equation in a body of mass and specific heat capacity c for a temperature change of$\mathit{\delta }\mathit{T}$ is given by $\mathit{Q}\mathbf{=}\mathit{m}\mathit{c}\mathit{\delta }\mathit{T}$.

Calculate the time using the heat flow equation

The rate of release of heat is $7\times {10}^{3}kJ/day=81.01J/s$.

Mass is , specific heat capacity is $3480J/k{g}^{0}C$and change in temperature is$0.{15}^{0}C$ .

Using the heat flow equation

$\begin{array}{r}\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}=\frac{mc\delta T}{t}\\ t=\frac{mc\delta T}{\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}\\ t=\frac{70\times 3480\times 0.15}{81.01}\\ t=456\mathrm{s}\\ t=7.6\min \end{array}$

So, the metabolism will take $7.6\min$to bring back the body to the initial temperature.