Question

A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be nearly the same temperature as the animal itself, so cooling by radiation is not effective. The only plausible way to keep cool is by evaporation, and elephants spray water on their body to accomplish this. If this is the only possible means of cooling, how many kilograms of water per hour must be evaporated from an elephant's skin to keep it at a constant temperature?

Answer 1

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

• That the temperature on the day is 35 °C
• That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

Answer 2

Answer:

$\frac{m}{t}$ $= 4.0$ $\frac{kg}{hr}$

Explanation:

$P=Q/s=(\frac{2500}{1}) (\frac{3600 s}{1 hr})=9$ × $10^{6} J/hr$

$Q = mL_{V} =$ $9$ × $10^{6} J$

→ $L_{V} = 22.6$ × $10^{5} Jkg$ (latent heat of vaporization of water)

∴ $P=\frac{mL_{V} }{t}$

Rearrange to find how many kilograms of water per hour is evaporated from the elephant's skin:

∴ $\frac{m}{t} = \frac{P}{L_{V}}$

$\frac{m}{t}=$ $9$ × $10^{6} J$ / $22.6$ × $10^{5}$

$\frac{m}{t}$ $= 4.0$ $\frac{kg}{hr}$