Question

A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1 000 J of energy from the higher-temperature reservoir and performs 250 J of work. Find (a) the entropy change of the Universe delta SU for this process and (b) the work W that could have been done by an ideal Carnot engine operating between these two reservoirs. (c) Show that the difference between the amounts of work done in parts (a) and (b) is T1 delta SU .

Answer 1

(a)
In steady state operation the net energy transfer to the engine is zero. That means heat delivered to the engine from hot reservoir equals work done plus heat rejected to cold reservoir
Q₂ = W + Q₁
=>
Q₁ = Q₂ - W = 1000 J - 750 J = 250 J


The total change in entropy of the universe equals the sum of the entropy changes of heat engine, cold reservoir(1) and hot reservoir (2). Since the heat engine operates in a cycle process is change in entropy is zero. Therefore,
∆Su = ∆S₁ + ∆S₂
Each reservoir exchanges heat at constant temperature. The change in entropy assigned to such a process is:
∆S =Q/T
(Note that Q is positive is heat is absorbed and negative if heat is rejected)
Hence
∆Su = (Q₁/T₁) + (Q₂/T₂)
= (750 J / 350K) + (- 1000 J / 600 K)
= 0.4762 J∙K⁻¹

(b)
The thermal efficiency of a heat engine, which is defined as the fraction of heat input which is converted to work, i.e.
η = W/Q₂
takes for an ideal Carnot engine the maximum value of:
η = 1 - (T₁/T₂)

Hence,
W = Q₂∙(1 - (T₁/T₂))
= 1000 J∙(1 - (350/600))
= 416.67.J

c)
∆W = W_real - W_ideal
= 416.67J - 250J
= 166.67 J

T₁∙∆Su = 350 K (0.4762 J∙K^{⁻ 1}) = 166.67 J
Q:E.D.