Question

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 heat of 2.26 x 106 Jlkg for the perspiration (same as for water), determine (a) the change in the player's internal energy, and (b) the number of nutritional calories the player has converted to work and heat.

Answer 1

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:

$\mathbf{dU = -491.6 \ kJ \times (\dfrac{1 \ cal}{ 4.186 \ J})}$

dU = -117.44 kcal

Learn more about first law of thermodynamics here:

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