Chapter 14, Problem 31P

(a)

To determine

The amount of heat leaving the body in a day only due to breathing.

Answer to Problem 31P

The amount of heat leaving the body in a day only due to breathing is $\underset{\_}{2\text{MJ}}$.

Explanation of Solution

Given that the volume of air breathed in one hour is $5{\text{m}}^3$, the pressure is $101\text{kPa}$, and the temperature is raised from $20\text{°C}$ to $35\text{°C}$.

Write the ideal gas law equation for the initial condition of the system.

$PV=nR{T}_{\text{i}}$ (I)

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T_{\text{i}}$ is the initial temperature.

Solve equation (I) for $n$.

$n=\frac{PV}{R{T}_{\text{i}}}$ (II)

Write the expression for heat energy corresponding to temperature change.

$Q=n{C}_{\text{p}}\Delta T$ (III)

Here, $Q$ is the heat, $C_{\text{p}}$ is the molar specific heat at constant pressure, and $\Delta T$ is the temperature change.

Use equation (II) in (III).

$Q=\frac{PV}{RT}{C}_{\text{p}}\Delta T$ (IV)

Write the expression for molar specific heat capacity of a diatomic gas.

$C_{\text{p}}=\frac{5}{2}R$ (V)

Use equation (V) in (IV).

$\begin{array}{c}Q=\frac{PV}{R{T}_{\text{i}}}\left(\frac{5}{2}R\right)\Delta T\\ =\frac{5PV\Delta T}{2T_{\text{i}}}\end{array}$ (VI)

Since the volume of air breathed in one hour is $5{\text{m}}^3$, in a whole day, the volume of air breathed is $\left(24\text{h}\times \text{5}{\text{m}}^3\text{/h}\right)$. The initial temperature is $20\text{°C}$ ($293.15\text{K}$)

Conclusion:

Substitute $101\text{kPa}$ for $P$, $\left(24\text{h}\times \text{5}{\text{m}}^3\text{/h}\right)$ for $V$, $\left(35\text{°C}-20\text{°C}\right)$ for $\Delta T$, and $293.15\text{K}$ for $T_{\text{i}}$ in equation (VI) to find $Q$.

$\begin{array}{c}Q=\frac{5\left(101\text{kPa}\right)\left(24\text{h}\times \text{5}{\text{m}}^3\text{/h}\right)\left(35\text{°C}-20\text{°C}\right)}{2\left(293.15\text{K}\right)}\\ =\frac{5\left(101\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}\right)\left(24\text{h}\times \text{5}{\text{m}}^3\text{/h}\right)\left(308.15\text{K}-293.15\text{K}\right)}{2\left(293.15\text{K}\right)}\\ =1.55\times {10}^6\text{J}\times \frac{1\text{MJ}}{1\times {10}^6\text{J}}\\ \approx 2\text{MJ}\end{array}$

Therefore, the amount of heat leaving the body in a day only due to breathing is $\underset{\_}{2\text{MJ}}$.

(b)

To determine

Whether the heat loss is significant, relative to the total heat loss from the body in one day.

Answer to Problem 31P

The heat loss is $\underset{\_}{\text{significant}}$ relative to the total heat loss from the body in one day and is about $\underset{\_}{20\%}$ of the total loss.

Explanation of Solution

Given that the total heat loss from the body in one day is $9\text{MJ}$. It is obtained that the amount of heat leaving the body in a day only due to breathing is $1.55\text{MJ}$.

Compute the fraction of the heat loss through breathing to the total heat loss.

$\frac{1.55\text{MJ}}{9\text{MJ}}\times 100\%=20\%$

Since the heat loss by breathing is about $20\%$ of the total heat loss, it is a significant amount.

Conclusion:

Therefore, the heat loss is $\underset{\_}{\text{significant}}$ relative to the total heat loss from the body in one day and is about $\underset{\_}{20\%}$ of the total loss.

(c)

To determine

The average power due to breathing alone in a day.

Answer to Problem 31P

The average power due to breathing alone in a day is $\underset{\_}{20\text{W}}$.

Explanation of Solution

Given that the time duration is $1\text{day}$. It is obtained that the heat loss through breathing is $1.55\text{MJ}$.

Write the expression for the power.

$\wp =\frac{Q}{\Delta t}$ (VII)

Here, $\wp$ is the power, $Q$ is the heat energy, and $\Delta t$ is the time duration.

Conclusion:

Substitute $1.55\text{MJ}$ for $Q$, and $1\text{day}$ for $\Delta t$ in equation (VII) to find $\wp$.

$\begin{array}{c}\wp =\frac{1.55\text{MJ}}{1\text{day}}\\ =\frac{1.55\text{MJ}\times \frac{1\times {10}^6\text{J}}{1\text{MJ}}}{1\text{day}\times \frac{24\text{h}}{1\text{day}}\times \frac{3600\text{s}}{1\text{h}}}\\ =18\text{W}\\ \approx 20\text{W}\end{array}$

Therefore, the average power due to breathing alone in a day is $\underset{\_}{20\text{W}}$.