91Ó°ÊÓ

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.