Question

Shivering is your body’s way of generating heatto restore its internal temperature to the normal $\mathbf{37}\mathbf{°}\mathbf{C}$ and it produces approximately 290 W of heat power per square meter of body area. A 68-kg, 1.78-m-tall woman has approximately1.8 m²of surface area. How long would this woman have to shiverto raise her body temperature by$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, assuming that the body loses none of this heat? The body’s specific heat capacity is about$\mathbf{3500}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{kg}\mathbf{·}\mathbf{K}$

Short answer

(a) The duration of shivering before the body temperature is raised by$1.0°\mathrm{C}$ is 4.6 min.

Step-by-step solution

Concept of Specific Heat.

Specific heat of a body is the amount of heat per unit mass required to elevate the temperature of one gram of the body by one degree Celsius.

Mathematically,

$\mathbf{Q}\mathbf{=}\mathbf{mc}\mathbf{∆}\mathbf{T}$

Here, m is the mass, c is the specific heat and$\mathbf{∆}\mathbf{T}$ is the change in temperature.

(a)Determination of the duration of shivering before the body temperature is raised by 1.0ºC.

Heat Power generated by the body is$290\mathrm{W}/{\mathrm{m}}^2$ and area given is $1.8{\mathrm{m}}^2$.

Total heat Power,

$$\begin{gathered} P=\left(290\mathrm{W}/{\mathrm{m}}^2\right)\times \left(1.8{\mathrm{m}}^2\right) \\ =522\mathrm{J}/\mathrm{s} \end{gathered}$$

Power expression is given as,

$$\begin{gathered} P=\frac{Q}{t} \\ t=\frac{Q}{P} \end{gathered}$$

From equation (i), the value of Q after substituting all the given values is,

$$\begin{gathered} Q=mc∆T \\ =\left(68\mathrm{kg}\right)\left(3500\mathrm{J}/\mathrm{kg}\right)\left(1.0°\mathrm{C}\right) \\ =2.38\times {10}^5\mathrm{J} \end{gathered}$$

Therefore, the required time is,

$$\begin{gathered} t=\frac{2.38\times {10}^5\mathrm{J}}{522\mathrm{J}/\mathrm{s}} \\ =456\mathrm{s} \\ =4.6\min \end{gathered}$$

The time for which women has to shiver is 4.6 min.