During a strenous workout, an athlete generates 1890.0 kj of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat?
To determine the mass of water that needs to evaporate from the athlete's skin to dissipate 1890.0 kJ of heat energy, we can use the equation:
Q = m * ΔHv
where:
Q is the heat energy in joules,
m is the mass of water in kilograms, and
ΔHv is the enthalpy of vaporization of water, which is 2.26 kJ/g.
First, let's convert the heat energy from kilojoules to joules. Since 1 kJ = 1000 J, 1890.0 kJ is equal to 1890.0 * 1000 = 1,890,000 J.
Next, we need to rearrange the equation to solve for m:
m = Q / ΔHv
Substituting the values, we get:
m = 1,890,000 J / (2.26 kJ/g * 1000 g/kg)
Simplifying further:
m = 1,890,000 J / 2,260 J/g
m ≈ 836.28 g
Therefore, approximately 836.28 grams (or 0.83628 kilograms) of water would need to evaporate from the athlete's skin to dissipate 1890.0 kJ of heat energy.