Question

1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the higher-temperature reservoir and performs 250 J of work. Find a. the entropy change of the Universe ΔSU 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 T1ΔSU.

Answer 1

Answer:

$\Delta S_u=2.1429\ J.K^{-1}$

$W_c=416.67\ J$

Explanation:

Given:

temperature of source reservoir, $T_H=600\ K$

temperature of sink reservoir, $T_L=350\ K$

energy absorbed from the source, $Q_{in}=1000\ J$

work done, $W=250\ J$

a.

Now change in entropy of the surrounding:

$\Delta S_u=\frac{dQ_L}{T_L}$

Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.

$\Delta S_u=\frac{Q_H-W}{T_L}$

$\Delta S_u=\frac{1000-250}{350}$

$\Delta S_u=2.1429\ J.K^{-1}$

b.

We know Carnot efficiency is given as:

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{350}{600}$

$\eta_c=0.4167$

Now the Carnot work done:

$W_c=Q_H\times \eta_c$

$W_c=1000\times 0.4167$

$W_c=416.67\ J$ .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

$\Delta W=W_c-W$

$\Delta W=416.67-250$

$\Delta W=166.67\ J$

Now, we find:

$T_L.\Delta S_u=350\times 2.1429$