Question

250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cools from ${60}^{\circ }C$ to ${55}^{\circ }C$ in 130 s and 67 s, respectively. If the water equivalent of the calorimeter is 10 g and rate of heat loss is same for both, then the specific heat of alcohol in ${\text{cal/g}}^{\circ }C$ is

• A. 1.3
• B. 0.67
• C. 0.62
• D. 0.985

Answer 1

The correct option is C 0.62

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
$=(250+10)5=260\times 5=1300\text{cal}$
Rate of loss of heat $=\frac{1300}{130}=10\text{cal/s}$
Heat lost by alcohol and calorimeter $=(200s+10)5$
Rate of loss of heat $=\frac{(200s+10)5}{67}\text{cal/s}$
Rate of loss of heat in the two cases are equal
$\therefore \frac{(200s+10)5}{67}=10$ or $s=0.62{\text{cal/g}}^{\circ }C$

Why this question? Caution: make sure to use specific heat of water in ${\text{cal/g}}^{\circ }C$ for the uniformity in units.