A) 81.5%
B) 26%
C) 40.5%
D) 51.5%
Correct Answer: B
Solution :
[b] Power sent to heat the water in the calorimeter \[P'=\frac{ms\Delta \theta }{t}=\frac{Vps\Delta \theta }{t}=\frac{{{10}^{3}}\times {{10}^{-6}}\times {{10}^{3}}\times 4200\times 4}{420}=40W\] Required ratio \[=\frac{P-P'}{P}=\frac{54-40}{54}=\frac{14}{54}=26\]%You need to login to perform this action.
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