Question

(1) An ideal Carnot engine operates between a hot reservoir at 500oC and a cold reservoir at 100oC with a heat input of 250 J per cycle from the hot reservoir.
(a) What is the efficiency of the Carnot engine?
(b) How much heat is delivered to the cold reservoir in each cycle?
(c) What is the change in entropy of the hot reservoir?
(d) What is the change in entropy of the cold reservoir?
(e) What minimum number of cycles is necessary for the engine to lift a 500 kg container of water through a height of 100 m?

Answer 1

necessaryAnswer:

a)$n=0,5175$

b)$Q_{c}=379.37J$

c)$dS_{h}=0.3234J/K$

d)$dS_{c}=1.017J/K$

e) 3787.6 cycles

Explanation:

As this engine is ideal, the efficiency depends only on the temperatura of the reservoirs(in kelvin scale):

$n=1-\frac{T_{c} }{T_{h}}=1-\frac{373}{773}=0.5175$

the efficiency is also related to the heat exchanged between the reservoirs:

$n=1-\frac{Q_{c} }{Q_{h}}$

Solving for $Q_{c}$ and having in mind that $Q_{h}=250J$:

$Q_{c}=(n+1)*Q_{h}=1.5175*250J=379.37J$

The change of entropy for the hot reservoir:

$dS_{h}=\frac{250J}{773K} =0.3234J/K$

Similar for the cold reservoir:

$dS_{c}=\frac{379.37J}{373K} =1.017J/K$

From the efficiency we can calculate the work that the engine can make for a cycle:

$n=\frac{W}{Q_{h}}$

solving for W:

$W=n*Q_{h}=0.5175*250J=129.37J$

$E=m*g*h=500kg*9.8m/s^{2} *100m=490000J$

It will take the engine:

$\frac{490000J}{129.37J/cycle}=3787.6cycles$