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$250\,gm$ of water and an equal volume of alcohol of mass $200\,gm$ are placed successively in the same calorimeter and cools from $60^{\circ}\,C$ to $55^{\circ}\,C$ in $130\,sec$ and $67 sec$ respectively. If the water equivalent of the calorimeter is $10\,gm$. , then the specific heat of alcohol in cal/gm $cal / gm ^{\circ}\,C$ is
$1.30$
$0.67$
$0.62$
$0.985$
Solution
(c)
Mass of water $=250 g$
Mass of alcohol $=200 g$
Water equivalent of calorimeter, $W=10 g$
Fall of temperature $=60-55=5^{\circ} C$
Time taken by water to cool $=130 s$
Time taken by alcohol to $cool =67 s$
Heat lost by water and colorimeter
$=(250+10) 5=260 \times 5=1300 cal$
Rate of loss of heat $=\frac{1300}{130}=10 cal / s$
Heat lost by alcohol and calorimeter $=(200 s+10) 5$
Rate of loss of heat $=\frac{(200 s +10) 5}{67} cal / s$
Heat lost by alcohol and calorimeter $=(200 s+10) 5$
Rate of loss of heat $=\frac{(200 s +10) 5}{67} cal / s$
Rates of loss of heat in the two cases are equal
$\frac{(200 s+10) 5}{67}=10$
$s=0.62\; cal / g^{\circ} C$