Question

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}{\mathit{m}}^{\mathbf{2}}$and the clothing is 1.0 cmthick; the skin surface temperature is $\mathbf{33}\mathbf{°}\mathit{C}$and the outer surface of the clothing is at$\mathbf{1}\mathbf{°}\mathit{C}$; the thermal conductivity of the clothing is$\mathbf{0}\mathbf{.}\mathbf{040}\mathbf{}\mathit{W}\mathbf{/}\mathit{m}\mathit{K}$.(b) If, after a fall, the skier’s clothes became soaked with water of thermal conductivity$\mathbf{0}\mathbf{.}\mathbf{60}\mathbf{}\mathit{W}\mathbf{/}\mathit{m}\mathbf{}\mathit{K}$, by how much is the rate of conduction multiplied?

Short answer

(a) The heat conducted through the clothing to the body is $230.4\mathrm{J}/\mathrm{s}$.

(b) The rate of conduction is multiplied by $15$.

Step-by-step solution

Stating the given data

i) Area on body is$A=1.8{\text{m}}^{\text{2}}$

ii) Thickness is$X=1.0 \text{cm or}0.01 \text{m}$

iii) Higher temperature is$T_f=33°Cor306K$

iv) Lower temperature is$T_i=1°Cor274K$

v) Thermal conductivity of the clothing,$k_{\text{clothing}}=0.040\text{W/mK}$

vi) Thermal conductivity of clothing soaked with water, $k_{\text{clothing}}=0.060\text{W/mK}$.

Understanding the concept of heat transfer

According to Fourier's law, the area at right angles to the gradient through which heat flows is directly proportional to the negative gradient of temperature and the time rate of heat transfer. We can find the total heat transferred and the rate of heat transfer using the formula for heat transfer rate.

Formula:

The rate of heat transfer via thermal energy per unit time is equal to the heat transfer and is given by

$Q=\frac{kA\Delta T}{X}$-----(i)

where k is thermal conductivity, $A$is area,$∆T$ is temperature change, and $X$is thickness.

(a) Calculation of heat conducted through clothing body

Now, using the formula of equation (i), we can get the heat transfer rateas follows:

$$\begin{gathered} Q=\frac{0.04 \text{W/mK}\times 1.8 {\text{m}}^{\text{2}} \times \left(306-274\right) \text{K}}{0.01 \text{m}} \\ =230.4 \text{J/s} \end{gathered}$$

Hence, the value of the conducted heat is localid="1663858113838" $230.4J/s$.

(b) Calculation of multiple factor of the conduction rate

Now, for the new value of thermal conductivity of clothing soaked with water, $k_{\text{clothing}}=0.060\text{W/mK}$, we get the heat transfer using equation (i) as

$$\begin{gathered} Q_1=\frac{0.6 \text{W/mK}\times 1.8 {\text{m}}^{\text{2}}\times \left(306-274\right) \text{K}}{0.01 \text{m}} \\ =3456\text{J/s} \end{gathered}$$

So, the factor that needs to be multiplied with the old value of heat transfer to get the above new value of heat transfer is

$$\begin{gathered} \frac{Q_1}{Q}=\frac{3456\text{J/s}}{230.4 \text{J/s}} \\ =15 \end{gathered}$$

Hence, the value of the multiplying factor is 15.