A runner and a cyclist are both training on a warm day they...

A runner and a cyclist are both training on a warm day. They can lose at most 1 kg of moisture per hour via sweating without becoming dehydrated. A person loses most of their excess thermal energy via the evaporation of sweat, and the latent heat of vaporization of sweat is 2430 kJ/kg. (a) What is the maximum rate of heat transfer from the bodies of each athlete? (b) The runner runs with an efficiency of 10%. What is the rate at which the runner can do work? (c) What is the change in internal energy of the runner over the course of 2 hours of training? (d) The cyclist cycles with an efficiency of 19%. Repeat parts (b) and (c) for the cyclist.

Transcript

00:01 Hello, here a runner and cyclist are boss training on a warm day.

00:04 So then they can lose almost one kilogram of liquid of moisture per hour.

00:17 The latin heat of vaporization is 2 ,430 kilojoules per kilogram.

00:26 And first, we have to calculate the maximum rate of heat transfer from the body of each athlete.

00:33 So this rate equals to the amount of transferred energy queue divided by time, which is 1 kilogram per hour times times 2 ,430 kilojoules per kilogram.

01:02 So that equals to 2430 kilojoules per hour.

01:07 And let's convert it to watts so therefore we have to divide it by 3600 seconds and multiply it by 1 ,000 so let's calculate it that equals to 675 watts so we've answered question a now the efficiency of runner is 10 % and we have to calculate the rate at which he can do the work, so we have to calculate delta w over delta.

02:00 So the efficiency in general is calculated as work divided by work plus absolute value of q.

02:16 So if you want to calculate delta v over delta t, we basically have to do the following substitution.

02:28 Or actually here it will be easier for us to first calculate work and then convert it to power.

02:36 Let's do this.

02:41 So here this fraction equals to 0 .10 and therefore work equals to 0 .10 w plus 0 .10 absolute value of q.

02:55 Therefore, 0 .90 work equals to 0 .10 q.

03:01 And now we can divide both parts by delta t therefore delta work over delta t equals to 0 equals to delta q or actually that equals to uh q or the rate which we already calculated divided by nine that equals to 675 watts divided by 9.

03:46 That equals to 75 watts...

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