A) 1.6
B) 1.2
C) 4.8
D) 3.5
Correct Answer: B
Solution :
The efficiency of heat engine is \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] |
or \[\frac{W}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] |
Here, \[{{Q}_{1}}=\] heat absorbed from the source of heat = 6 kcal |
\[{{T}_{1}}=\] temperature of source |
\[=227+273=500\text{ }K\] |
and \[{{T}_{2}}=\] temperature of sink |
\[=127+273=400\text{ }K\] |
Hence, \[\frac{W}{6}=1-\frac{400}{500}\] |
or \[\frac{W}{6}=\frac{100}{500}\] |
or \[W=1.2\text{ kcal}\] |
Thus, amount of heat converted into work is 1.2 kcal. |
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