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 K and \[{{T}_{2}}=\] temperature of sink = 127 + 273 = 400 K Hence, \[\frac{W}{6}=1-\frac{400}{500}\] or \[\frac{W}{6}=\frac{100}{500}\] or W = 1.2 kcal Thus, amount of heat converted into work is 1.2 kcal.You need to login to perform this action.
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