Question

Temperature of source and sink of carnot engine are 400k and 300k its efficiency will be

Answer 1

η = 1 - T1/T2

T1=300 K ; T2=400 K

Hence,

η = 1 - 300/400

=>η = 1 - 0.75

Hence,

η = 0.25

= 25%
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Answer 2

The efficiency of the Carnot engine is 25 %.

Explanation:

It is given that,

Temperature of the source, $T_1=400\ K$

Temperature of the sink, $T_2=300\ K$

To find,

Efficiency of the Carnot engine

Solution,

The efficiency of the Carnot engine in terms of temperature of sink and the source is given by :

$\eta=1-\dfrac{T_2}{T_1}$

$\eta=1-\dfrac{300}{400}$

$\eta=0.25$

or

$\eta=25\%$

So, the efficiency of the Carnot engine is 25%. Therefore, it is the required solution.

Learn more,

Carnot engine

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