Question

$250g$ of water and equal volume of alcohol of mass $200g$ are replaced successively in the same calorimeter and cool from $606^{\circ }C$ to $55^{\circ }C$ in $130s$ and $67s$, respectively. If the water equivalent of the calorimeter is $10g$, 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 $=250g$
Mass of alcohol $=200g$
Water equivalent of calorimeter, $W=10g$
Fall of temperature $=60-55=5^{\circ }C$
Time taken by water to cool $=130s$
Time taken by alcohol to cool $=67s$
Heat lost by water and calorimeter
$=(250+10)5=260\times 5=1300cal$
Rate of loss of heat $=\frac{1300}{130}=10cal/s$
Heat lost by alcohol and calorimeter $=(200s+10)5$
Rate of loss of heat $=\frac{(200s+10)5}{67}cal/s$
Rates of loss of heat in the two cases are equal
$\therefore \frac{(200s+10)s}{67}=10$
or $s=0.62cal/g^{\circ }C$