Sweaty Biker and Energy Consumption

[yoshiko]

In a hot day’s race, a bicyclist consumes 8L water over 4 hrs.
Assume that 80% of his energy goes into evaporating this water as sweat. (This isn't a bad approximation, b/c the mechanical efficiency of a bicycle rider is only ~20%, w/ the rest of the energy consumed going into heat.)
How much total energy, in kcal, did the rider use during the race?

I know that 8L is evaporated by heat (body). I can obtain the mass of water, giving me Q.

I need to use the equation Lxm= Q, from this get 80% of total --> total
Latent heat and vaporization

You can get the mass from knowing 1L= 1000cm^3
Because there are 8L, you have 8000cm^3
Convert to grams knowing the density of water: 1g/cm^3
(8000cm^3)(1g/cm^3)= 8000g or 8 kg water.

I am stuck from here please help!

 

[Shooting Star]

Energy to convert 8 Litres of water into vapour = mass*latent heat = (8000 gm)*(540 cal/gm). This is 80% of total energy. What's the problem now?

 

[ogb p]

To calculate the total energy used by the rider, we need to use the formula for latent heat of vaporization (L) and the mass of water (m). Latent heat of vaporization is the energy required to change a substance from liquid to gas without changing its temperature.

The formula for calculating latent heat of vaporization is Q = L x m.

We already know the mass of water (8 kg) and we can find the latent heat of vaporization of water from a table or by looking it up online. The value is approximately 2,260,000 joules per kilogram (J/kg).

Now, we need to convert the units of energy from joules to kilocalories (kcal). 1 kcal is equal to 4184 joules.

So, total energy used by the rider = Q x (80%) / 4184 = (2,260,000 J/kg x 8 kg x 0.80) / 4184 = 3,431.2 kcal

Therefore, the rider used approximately 3,431.2 kcal of energy during the race. It is important to note that this is just an estimate and the actual energy consumption can vary based on factors such as the rider's weight, intensity of the race, and environmental conditions.