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, \[{{H}_{1}}=m{{L}_{v}}=540\,\,m\,cal\] |
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.
You will be redirected in
3 sec