Question

$250\, g$ of water and equal volume of alcohol of mass $200 \, g$ are replaced successively in the same calorimeter and cool from $606^{\circ} C$ to $55^{\circ} C$ in $130\, s$ and $67\, s$, respectively. If the water equivalent of the calorimeter is $10 \, g$, then the specific heat of alcohol in $cal / g ^{\circ} C$ is

• A. $1.30$
• B. $0.67$
• C. $0.62$
• D. $0.985$

Answer 1

Answer: (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 calorimeter
$=(250+10) 5=260 \times 5=1300\, cal$
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 $=\frac{(200 s+10) 5}{67} cal / s$
Rates of loss of heat in the two cases are equal
$\therefore \frac{(200 s+10) s}{67}=10 $
or $ s=0.62 \,cal / g ^{\circ} C$