Question

Imagine you have a hot reservoir at a temperature of 89.0 °C, and cold reservoir at a temperature of 15.0 °C. Given their vast size, it is reasonable to assume the reservoirs\' temperatures will not change significantly if heat flows into or out of them. These reservoirs are then brought into thermal contact, during which 39210 J of heat flows from the hot reservoir to the cold reservoir. As a result of this heat exchange, what is the total change in entropy? What is the most work that could be done by a heat engine if it takes in the 39210 J from the hot reservoir and exhausts it into the cold reservoir?

Answer 1

the total change in entropy = Q/T

                                           = 39210/(273.15 + 89)

                                          = 108.27 J/K

most work that could be done by a heat engine = 39210 * [(89 - 15)/(273.15 + 89)]

                                                                            = 8012 J

Answer 2

SOLUTION :

Total change in entropy ( ∆S ) 

= H received / T receiver  - H given / T giver

= 39210 / (273+15)  - 39210 / (273+89)

= 39210 / 288 - 39210 / 362

= 27.83 J/K (ANSWER)

Most work done by the Heat Engine

= Heat taken by the engine * ( 1 - T cold / T hot)

= 39210 * (1 - 288 / 362) 

= 8015.30 J (ANSWER).