00:02
Hello students, first of all in the first part the maximum rate of heat transfer from each athlete is equals to 3 the rate at which they lose the moisture via sweating which is 1 kg per hour.
00:22
The latent heat l of vaporization of sweat is 2 .23 into 10 to the power 6 joule per kg.
00:32
So, maximum rate of heat transfer is equals to 1 kg per hour into 2 .23 into 10 to the power 6 joule per kg which is equals to 2 .23 into 10 to the power 6 joule per hour.
01:01
Next the runner's efficiency is given to be 10 percent which means that only 10 percent of the energy the use is converted into the work the less the rest energy is lost as heat.
01:24
So, the rate at which runner do work is equals to 0 .1 into 2 .23 into 10 to the power 6 which is equals to 2 .23 into 10 to the power 5 joule per hour.
01:48
In the next part it says to calculate the change in internal energy.
01:54
The change in internal energy for that first of all the total energy they use is given by total energy is equals to 2 .23 into 10 to the power 5 joule per hour into the time that is 3 hours.
02:20
So, the total energy is 6 .69 into 10 to the power 5 joule.
02:26
The energy they lose as heat energy lost as heat is equals to 1 into 2 .23 into 10 to the power 6 into 3 as it is for 3 hours which is equals to 6 .69 into 10 to the power 6 joule...