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Olenka [21]
3 years ago
6

(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 pe

r 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?
Physics
1 answer:
Vlad [161]3 years ago
3 0

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

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