Question

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temperature of 378 K. Over the course of one hour, the engine absorbs 1.42 × 10^5 J from the hot reservoir and exhausts 7.5 × 10^4 J into the cold reservoir.

a. What is the power output of this engine?
b. What is the maximum (Carnot) efficiency of a heat engine running between these two reservoirs?
c. What is the actual efficiency of this engine?

Answer 1

Answer:

a. $P=18.61\ W$

b. $\eta_c=0.6976=69.76\%$

c. $\eta=0.4718=47.18\%$

Explanation:

Given:

• temperature of the hotter reservoir, $T_H=1250\ K$

• temperature of the colder reservoir, $T_C=378\ K$

• heat absorbed by the engine, $Q_H=1.42\times 10^{5}\ J$

• heat rejected to the cold reservoir, $Q_L=7.5\times 10^4\ J$

• time duration of the energy transfer, $t=1\ hr=3600\ s$

Now the work done by the engine:

Using energy conservation,

$W=Q_H-Q_L$

$W=14.2\times 10^4-7.5\times 10^4$

$W=6.7\times 10^4\ J$

a.

Hence the power output:

$P=\frac{W}{t}$

$P=\frac{6.7\times 10^4}{3600}$

$P=18.61\ W$

b.

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

$\eta_c=1-\frac{378}{1250}$

$\eta_c=0.6976=69.76\%$

c.

now actual efficiency:

$\eta=\frac{W}{Q_H}$

$\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}$

$\eta=0.4718=47.18\%$