A) 0.098°C
B) 0.98°C
C) 9.8°C
D) 0.0098°C
Correct Answer: A
Solution :
As \[W=JQ\] Þ \[\frac{1}{2}(mgh)=J\times mc\Delta \theta \] Þ \[\Delta \theta =\frac{gh}{2JS}\] \[\Delta \theta =\frac{9.8\times 84}{2\times 4.2\times 1000}=0.098{}^\circ C\] \[(\because \,{{S}_{\text{water}}}=1000\frac{cal}{kg\times {}^\circ C})\] Short trick: Remember the value of\[\frac{g}{J{{c}_{W}}}=0.0023\], here \[\Delta \theta =\frac{1}{2}\times (0.0023)h=\frac{1}{2}\times 0.0023\times 84=0.098{}^\circ C\]
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