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An heat engine absorbs heat at 327oC and exhausts heat at 127oC. The maximum amount of work performed by the engine in joule per kilo calorie is X. Find X5.

Answer
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Hint: For solving this question we have to consider the concepts of heat engine, we have to use temperature in the standard unit of kelvin here. With the help of efficiency formula and work done by heat engine we will determine the value of X. Mainly, one must calculate efficiency and work done using efficiency.

Formula used:
1. η=1T2T1
Where, η is the efficiency of the heat engine, T1 is the temperature at which it absorbs and T2 is the temperature at which it exhausts.
2. W=Qη
Where, Wis work done by the heat engine and Qis the source heat in kilo calorie.

Complete answer:
Let us begin with the conversion of the temperature into standard units such as
T1=327oC=(327+273)K=600K
T2=127oC=(127+273)K=400K

So, let’s begin with calculating the efficiency of the heat engine, we have
η=1T2T1

Let us substitute all the given values in the formula above, we get
η=1400K600K
η=123=10.66=0.34
η=0.34

Now, we have to find work for per kilo calorie, for that we have a formula for work done as below:
W=Qη
But, the source heat is Q=H=1kcal
Also, remember that
1kcal=4.2×103joule

Let us put all these values in the formula for work done for calculating total work.
W=4.2×103joule×0.34
W=1.428×103joule

But here work done is given in the form of X. Therefore,
X=W=1.428×103joule
X=1428J

Therefore, X5 is given by:
X5=1428J5=285.6J

So, the answer is 285.6J.

Note: Here, the question is designed in such a way that there is just one concept that has been used to work efficiently. We have to recall all the important points from the heat engine and apply it over here.