A basketball player does 2.43 x 105 J of work during her time in the game, and evaporates 0,1 '10 kg of water. Assuming a latent
1 answer:
The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal
For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.
The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.
From the given information,
- the work done on the basketball is dW = 2.43 × 10⁵ J
The amount of heat loss is represented by dQ.
where;
- dQ = -mL
∴
Using the first law of thermodynamics:b
dU = dQ - dW
dU = -mL - dW
dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)
dU = -491.6 × 10³ J
dU = -491.6 kJ
The number of nutritional calories the player has converted to work and heat can be determined by using the relation:
dU = -117.44 kcal
Learn more about first law of thermodynamics here:
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