250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from 606^oC to 55^oC 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-^oC is

Mass of water = 250 g
Mass of alcohol = 200 g
Water equivalent of calorimeter, W = 10 g
Fall of temperature = 60 - 55 = 5^oC
Time taken by water to cool = 130 s
Time taken by alcohol to cool = 67 s
Heat lost by water and calorimeter

                                       $=(250+10)5=260\times 5=1300\mathrm{cal}$

Rate of loss of heat $=\frac{1300}{130}=10\mathrm{cal}/\mathrm{s}$

Heat lost by alcohol and calorimeter $=(200s+10)5$

Rate of loss of heat $=\frac{(200s+10)5}{67}\mathrm{cal}/\mathrm{s}$

Rates of loss of heat in the two cases are equal

$\therefore \frac{(200s+10)5}{67}=10\text{ or }s=0.62\mathrm{cal}/{\mathrm{g}}^{\circ }\mathrm{C}$