A) 100
B) 54
C) 3.6
D) 8.2
Correct Answer: C
Solution :
Let m is the amount of steam that passes off the radiator. Heat released due to condensation of steam, \[{{H}_{1}}=m{{L}_{v}}=540mcal\] Heat released due to cooling of water, \[{{H}_{2}}=ms\Delta \theta =m\times 1\times 20=20m\,cal\] Required percentage \[=\frac{{{H}_{2}}}{{{H}_{1}}+{{H}_{2}}}\times 100=\frac{20m}{560m}\times 100=3.6%\]You need to login to perform this action.
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