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Evaporating sweat cools the body because evaporation is an endothermic process. Estimate the mass of water that must evaporate from the skin to cool the body by 1.50°C. Assume a body mass of 84.0 kg and assume that the specific heat capacity of the body is 4.00 J/g°C.

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Final answer:

Approximately 9000 joules of heat is required to evaporate 1.5 kg of water and cool the body by 1.50°C.

Step-by-step explanation:

Since evaporation is an endothermic process, perspiration evaporating cools the body. The following formula can be used to determine the quantity of water that must evaporate from the skin in order to chill the body by 1.50°C:

Q = mcΔT

Where

Q is the amount of heat required,

m is the mass of water,

c is the specific heat capacity, and

ΔT is the change in temperature.

Assuming that perspiration is primarily composed of water, the estimated mass of water in this instance is 1.5 kg, given the body mass of 84.0 kg. The body has a 4.00 J/g°C specific heat capacity, or 4000 J/kg°C equivalent. There has been a 1.50°C temperature shift.

Substituting the values into the formula:

Q = (1.5 kg)(4000 J/kg°C)(1.50°C) = 9000 J

Therefore, approximately 9000 Joules of heat are required for 1.5 kg of water to evaporate and cool the body by 1.50°C.

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