A basketball player does 2.22×10^5 J of work during her time in the game, and 0.150 kg of water evaporates from her skin. Assume a latent heat of 2.26×10^6J/kg for the evaporation of sweat (the same as for water). Determine the change in the player’s thermal energy
To determine the change in the player's thermal energy, we need to first calculate the amount of heat gained by the player through evaporation.
The formula to calculate the amount of heat gained or lost during a phase change (such as evaporation) is:
Q = m * L
Where:
Q is the amount of heat gained or lost
m is the mass of the substance
L is the latent heat of the substance
In this case, we are given the mass of water evaporated (0.150 kg) and the latent heat of evaporation (2.26×10^6 J/kg). Therefore, we can calculate the amount of heat gained during evaporation as follows:
Q = m * L
Q = 0.150 kg * 2.26×10^6 J/kg
Q = 3.39×10^5 J
So, the amount of heat gained by the player through evaporation is 3.39×10^5 J.
Since the player did work during the game and work is a form of energy, the change in the player's thermal energy can be calculated by subtracting the work done from the heat gained through evaporation:
Change in thermal energy = Heat gained - Work done
Change in thermal energy = (3.39×10^5 J) - (2.22×10^5 J)
Change in thermal energy = 1.17×10^5 J
Therefore, the change in the player's thermal energy is 1.17×10^5 J.