Chapter 3: Q50E (page 577)

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0 kg man to cool his body $1 . 00^{0} C$ ? The heat of vaporization of water at body temperature ( $37^{0} C$ ) is . The specific heat of a typical human body is 3480J/kgK (see Exercise 17.25 ). (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can ( $355 c m^{3}$ ).

Short Answer

Expert verified

a) 0.101kg

b) 101cm

Step by step solution

Equate the heat of evaporation with change in body temperature

Given that $m_{m a n} = 70 k g, δ T = 1^{0} C, T_{1} = 37^{0} C, a n d L_{v} = 2.42 \times 10^{6} J / k g$ and (for water).

For the evaporation of the water, the heat flow equation is $Q = m L_{v}$ .

For the change in body temperature, the heat flow equation is $Q = m c δ T$ .

Equate the heat flow equations.

$- m_{b o d y} c_{b o d y} δ T = m_{w a t e r} L_{v m c} m_{w} = - \frac{m_{b o d y} \times c_{b o d y} \times δ T}{L_{v}} m_{w} = - \frac{70 \times 3480 \times (- 1)}{242 \times 10^{4}} m_{w} = 0.101 k g$

Calculate the volume of water using the density of water

The density of water is $1000 k g / m^{3}$ .

$V = \frac{m_{w}}{ρ} V = \frac{0.101}{1000} V = 0.101 \times 10^{- 3} m^{3} V = 101 c m^{3}$

By comparing the water volume result with soft drinking $\frac{101}{355} \times 100 = 35 %$ .

The volume of water needed to replenish the evaporated water is less than soft drink with 65%.

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Step by step solution

Equate the heat of evaporation with change in body temperature

Calculate the volume of water using the density of water

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