Question

Evaporating sweat cools the body because evaporation is an endothermic process:

H2O(l)→H2O(g)ΔH∘rxn=+44.01kJ

Estimate the mass of water that must evaporate from the skin to cool the body by 0.55 ∘C∘ Assume a body mass of 95 kg and assume that the specific heat capacity of the body is 4.0 J/g⋅∘C

Express your answer using two significant figures.

Answer 1

The mass of water that must evaporate from the skin to cool the body by 0.55 °C is approximately 2.0 kg.

Step-by-step explanation:

According to the information given, the enthalpy change for the evaporation of water is +44.01 kJ/mol. To estimate the mass of water that must evaporate from the skin to cool the body by 0.55 °C, we can use the formula:

mass of water = (heat capacity of body * body mass * temperature change) / enthalpy of vaporization

Plugging in the values, we have:

mass of water = (4.0 J/g⋅°C * 95 kg * 0.55 °C) / (44.01 kJ/mol * 1000 g/kg)

Solving this, we find that the mass of water that must evaporate is approximately 2.0 kg.

Answer 2

Approximately 85.6 grams of water need to evaporate from the skin to cool the body by 0.55°C.

How can you estimate the mass of water that must evaporate from the skin to cool the body by 0.55 ∘C?

We know the body mass (m) is 95 kg and the temperature change (ΔT) is 0.55 °C.

We need the specific heat capacity (c) of the body, which you provided as 4.0 J/g°C. We need to convert the body mass to grams for consistency: 95 kg * 1000 g/kg = 95000 g.

Now we can use the formula: Q = mcΔT, where Q is the heat required. Plugging in the values, we get: Q = 95000 g * 4.0 J/g°C * 0.55 °C = 209000 J.

We know the enthalpy change (ΔH) for water evaporation is 44.01 kJ/mol (note that 1 kJ is equal to 1000 J).

We need to convert the heat required (Q) to kJ: 209000 J / 1000 J/kJ = 209 kJ.

Now we can calculate the mass of water (m_water) needed to evaporate using the formula: m_water = Q / ΔH. Plugging in the values, we get: m_water = 209 kJ / 44.01 kJ/mol = 4.75 mol.

Finally, we need to convert the moles of water to grams. The molar mass of water is 18.02 g/mol. So, the mass of water needed is: 4.75 mol * 18.02 g/mol = 85.57 g.

Therefore, approximately 85.6 grams of water need to evaporate from the skin to cool the body by 0.55°C. This is just an estimate, and the actual amount may vary depending on factors like humidity, air temperature, and individual physiology.