Question

a basketball player does 2.43×10^5 joules of work during her time in the game and evaporates 0.110 kilograms of water through perspiration. assume the latent heat of vaporization for water is 2.26×10^6 joules per kilogram. what is the change in the players internal energy?

a. -492 kj
b. 362 kj
c. 154 kj
d. 916 kj
e. 751 kj

Explanation:

Step1: Calculate heat lost due to evaporation

The heat lost due to evaporation $Q$ is given by the formula $Q = mL$, where $m$ is the mass of water evaporated and $L$ is the latent - heat of vaporization. Given $m = 0.110\ kg$ and $L=2.26\times 10^{6}\ J/kg$. So, $Q=(0.110\ kg)\times(2.26\times 10^{6}\ J/kg)=2.486\times 10^{5}\ J$.

Step2: Use the first - law of thermodynamics

The first - law of thermodynamics is $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system, and $W$ is the work done by the system. Here, the work done by the player $W = 2.43\times 10^{5}\ J$ and the heat lost (negative heat added to the player) $Q=- 2.486\times 10^{5}\ J$. Then $\Delta U=-2.486\times 10^{5}\ J-2.43\times 10^{5}\ J=-4.916\times 10^{5}\ J=-491.6\ kJ\approx - 492\ kJ$.

Answer:

A. -492 kJ