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Naya [18.7K]
3 years ago
7

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.
Chemistry
1 answer:
8_murik_8 [283]3 years ago
4 0

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

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The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

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Mg(ClO3)2 —> MgCl2 + O2

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Mg(ClO3)2 —> MgCl2 + 3O2

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The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

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There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

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C5H12 + O2 —> 5CO2 + 6H2O

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C5H12 + 8O2 —> 5CO2 + 6H2O

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