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.

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.

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 =

2b2t

explanation:

2b2t

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2