Question

Q4: The average heat evolved by the oxidation of food in an average adult per hour per kilogram of body weight is 7.20 ୩୎ ୩୥ ୦୰. Assume the weight of an average adult is 62.0 kkg. Suppose the total heat evolved by this oxidation is tranderred into the surroundings over a period of 1 week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at 293K.

Answer 1

Answer:

The entropy change of the surroundings associated with this heat transfer is 255.9J/K

Explanation:

The heat evolved by the oxidation of food is 7.2J/Kg.h thus, multiplying the heat by the weight of the person and by the time the total heat (q) is obtained. In this case, the weight is 62.0Kg and a weak has 168 hours.

$q = 7.2\frac{J}{Kg.h}x62.0Kgx168h = 74995.2J$

The entropy change of the surrounding associated with this heat transfer can be calculated by the ratio between the heat exchanged and the temperature.

ΔS = $\frac{q}{T} = \frac{74995.2J}{293K} = 255.9J/K$