A student eats 2000 kcal per day. (a) Assuming that all of the food energy is released as heat, what is the rate of heat released (in watts)? (b) What is the rate of change of entropy of the surroundings if all of the heat is released into air at room temperature \(\left(20^{\circ} \mathrm{C}\right) ?\)

First, we need to convert the 2000 kcal into joules. Recall that 1 kcal = 4184 J. So we have: 2000 kcal/day × (4184 J/kcal) = 8,368,000 J/day

To find the rate of heat released in watts, we need to convert joules to watts. Recall that power (W) is energy per unit time (J/s), and we are given energy per day. Since 1 day = 86400 seconds, we can use the following conversion: 8,368,000 J/day × (1 day/86400 s) = 96.85185 W

We are given the room temperature as 20°C, which we need to convert to Kelvin. The Kelvin temperature can be found by adding 273.15 to the Celsius temperature: \(T = 20 + 273.15 = 293.15 K\) Now, we can use the following formula to find the rate of change of entropy: \(\frac{dS}{dt} = \frac{P}{T}\) where \(\frac{dS}{dt}\) is the rate of change of entropy, \(P\) is the power in watts, and \(T\) is the temperature in Kelvin. Substitute our calculated values into the formula: \(\frac{dS}{dt} = \frac{96.85185 W}{293.15 K} = 0.3306 \frac{W}{K}\) #Conclusion# The rate of heat released as the student consumes 2000 kcal per day is approximately 96.85 W, and the rate of change of entropy of the surroundings is approximately 0.3306 W/K.