Question

A bicyclist consumes 9.0 L of water over the span of 3.5 hours during a race. Making the approximation that 80% of the cyclist’s energy goes into evaporating this water (at 20°C) as sweat, how much energy in kcal did the rider use during the ride? (Hint: See page 399.)

Short answer

The rider uses \(6.6 \times {10^3}\;{\rm{kcal}}\) energy during the ride.

Step-by-step solution

Given data

Water consumed by the cyclist is\(V = 90\;{\rm{L}}\).

Let the rider use Q energy during the ride.

Assume that the total consumed water gets vaporized.

Concepts

The mass of 1 L water is 1 kg.

The heat absorbed by the water to evaporate is equal to the 80% energy of the cyclist.

Calculation of part (a)

The mass of 9.0 L water is \(m = 9\;{\rm{kg}}\).

The latent heat of vaporization is \(L = 585\;{\rm{kcal/kg}}\).

\(\begin{aligned}{c}\frac{{80}}{{100}}Q = mL\\Q = \frac{{100}}{{80}}mL\\Q = \frac{{100}}{{80}} \times \left( {9\;{\rm{kg}}} \right) \times \left( {585\;{\rm{kcal/kg}}} \right)\\Q = 6.6 \times {10^3}\;{\rm{kcal}}\end{aligned}\)

Hence, the rider uses \(6.6 \times {10^3}\;{\rm{kcal}}\) energy during the ride.