Question

When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir

(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?

Answer 1

a)
`\Delta S_(in) = 2.791\,(J)/(K)`, b)
`\Delta S_(out) = 6.897\,(J)/(K)`, c)
`S_(gen) = 4.106\,(J)/(K)`, d) Due to irreversibilities due to temperature differences.

Step-by-step explanation:

a) The change in entropy of the hot reservoir is:

`\Delta S_(in) = (2400\,J)/(860\,K)`

`\Delta S_(in) = 2.791\,(J)/(K)`

b) The change in entropy of the cold reservoir is:

`\Delta S_(out) = (2400\,J)/(348\,K)`

`\Delta S_(out) = 6.897\,(J)/(K)`

c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:

`\Delta S_(in) - \Delta S_(out) + S_(gen) = 0`

`S_(gen) = \Delta S_(out) - \Delta S_(in)`

`S_(gen) = 6.897\,(J)/(K) - 2.791\,(J)/(K)`

`S_(gen) = 4.106\,(J)/(K)`

d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.