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Mice21 [21]
3 years ago
9

A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

maintained at 210 K. Find the temperature of the heat source.
Physics
1 answer:
bixtya [17]3 years ago
6 0

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

\eta = 1 - \frac{T_2}{T_1}

Where,

T_2 = Temperature of Cold Body

T_1 =Temperature of Hot Body

\eta= Efficiency

According to the statement our values are:

\eta = 0.3

T_2 = 210K

Replacing we have that

0.3 = 1- \frac{210}{T_1}

\frac{210}{T_1} = 0.7

T_1 = \frac{210}{0.7}

T_1 = 300K

Therefore the temperature of the heat source is 300K

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s_2=v\times t_2

s_2=12.8\times t_1\times t_2

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