Question

In one cycle, a heat engine absorbs 520 J from a high-temperature reservoir and expels 310 J to a low-temperature reservoir. If the efficiency of this engine is 59% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?

Answer 1

Work done by the engine per cycle
W = Qh - Qc = 520J - 310J = 210J

efficiency of the engine
? = W/Qh = 210J/520J = 0.403

The engine operates at 59% of maximum attainable efficiency (carnot efficiency), i.e.
? = 0.59??c

Hence:
?c = ?/0.59 = 0.403 / 0.59 = 0.683

Carnot efficiency is related to the cold temperature to hot temperature ratio as:
?c = 1 - Tc/Th

Hence:
Tc/Th = 1 - ?c = 1 - 0.683 = 0.317

Answer 2

1)
the actual efficiency of the engine is:
? = W/Qhigh = (Qhigh - Qlow)/Qhigh = 1 - Qlow/Qhigh

The ideal Carnot efficiency would be
?c = 1 - Tlow/Thigh

Moreover you know that operates at 59% of Carnot efficiency.
Hence:
? = 0.59??c
<=>
1 - Qlow/Qhigh = 0.59?(1 - Tlow/Thigh)
<=>
Tlow/Thigh = (Qlow/Qhigh - 0.45)/ 0.59
= (310J/520J - 0.41)/ 0.59
= 0.3155

Answer 3

W = Qh - Qc = 520J - 310J= 210J

efficiency of the engine
? = W/Qh = 210J/520J = 0.4038

The engine operates at 59% of maximum attainable efficiency (carnot efficiency), i.e.
? = 0.59??c

Hence:
?c = ?/0.59 = 0.4038/ 0.59 = 0.6844

Carnot efficiency is related to the cold temperature to hot temperature ratio as:
?c = 1 - Tc/Th

Hence:
Tc/Th = 1 - ?c = 1 - 0.6844 = 0.3155

Answer 4

Efficiency of engine,

$\eta=1-\frac{Q_L}{Q_H}$

$\Rightarrow \eta=1-\frac{310}{520}=0.4038$

Efficiency of Carnot Cycle,

$\eta_C=\frac{\eta}{0.59}=\frac{0.4038}{0.59}=0.685$

We know that,

$\eta_C=1-\frac{T_L}{T_H}=0.684$

$\Rightarrow \frac{T_L}{T_H}=1-0.684=0.316$