Question

Penguins huddling: To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups. Assume that a penguin is a circular cylinder with a top surface area a =0.26 $m^2$ and height h = 90 cm. Let $P_r$ be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus $N{P}_r$ is the rate at which N identical, well-separated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area Na and height h, the cylinder radiates at the rate $P_h$. If N = 1000, by what percentage does huddling reduce the total radiation loss?

• A. 33$\%$
• B. 52$\%$
• C. 13$\%$
• D. 27$\%$

Answer 1

The correct option is D

27$\%$

Heat loss by individula penguin =$\sigma e{A}_{penguin}(T_{penguin}^4-T_{environment}^4)$

=$\sigma e(2\pi r_{penguin}h+2\pi r_{penguin}^2)(T_{penguin}^4-T_{environment}^4)$

Let, $\sigma e(T_{penguin}^4-T_{environment}^4)=k$

$\therefore Heatlostby1000penguinsradiatingfarawayfromeachother,Q_1$

$=1000(2\pi r_{penguin}h+2\pi r_{penguin}^2)k$

Now, $\pi r_{penguin}^2=0.26m^{2}or{r}_{penguin}=0.287m$

Also, $\pi r_{huddle}^2=N{a}_{penguin}=1000\times 0.26or{r}_{huddle}=9.09m$

$\therefore Heat$ lost by hudle of [enguins,$Q_2$

$=\sigma e{A}_{huddle}(T_{penguin}^4-T_{environment}^4)=(2pi{r}_{huddle}h+2\pi r_{huddle}^2)k$

NOw, percentage of reduce in radiation loss is given by $\frac{Q_2}{Q_1}=(2pi{r}_{huddle}h+2\pi r_{huddle}^2)\times 100/$

$N(2pi{r}_{penguin}h+2\pi r_{huddle}^2)=26.67\%$