Questions: During a strenuous workout, an athlete generates 1940 kJ of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate 1940 kJ of heat?

Solution Steps

Since the problem involves evaporation of water from the athlete's skin, which is likely at approximately room temperature, use the molar heat of vaporization at \(25.0^{\circ} \mathrm{C}\), which is \(44.0 \mathrm{~kJ} / \mathrm{mol}\).

Use the formula: \[ \text{Moles of water} = \frac{\text{Total heat energy}}{\text{Molar heat of vaporization}} \] Substitute the given values: \[ \text{Moles of water} = \frac{1940 \, \text{kJ}}{44.0 \, \text{kJ/mol}} \]

Convert moles of water to mass using the molar mass of water (\(18.015 \, \text{g/mol}\)): \[ \text{Mass of water} = \text{Moles of water} \times 18.015 \, \text{g/mol} \]

Final Answer