Question

An engineer claims to have measured the characteristics of a heat engine that takes in 150 J of thermal energy and produces 50 J of useful work. What is the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs?

Answer 1

Answer:

1.4999

Explanation:

Efficiency can be calculated using below expresion

Efficiency = W/Q.............eqn(1)

Where W= work = 50 J

Q= thermal energy= 150 J

But

W/Q= (Th-Tc)/Th ...........Eqn(2)

Th= temperature of the hot

Tc= temperature of the cold

Where Th/ Tc= ratio of the temperature hot and cold reservoirs?

If we simplify eqn(2) we have

W/Q = 1-Tc/Th.........eqn(3)

If we make the ratio subject of the formula we have

Tc/Th = 1-(W/Q)

Th/Tc = 1/(1-W/Q )

Then substitute the values

= 1/(1-50/150) = 1.4999

Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 1.4999