250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from 606oC to 55oC 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 gMass of alcohol = 200 gWater equivalent of calorimeter, W = 10 gFall of temperature = 60 - 55 = 5oCTime taken by water to cool = 130 sTime taken by alcohol to cool = 67 sHeat lost by water and calorimeter                                       =(250+10)5=260×5=1300calRate of loss of heat =1300130=10cal/sHeat lost by alcohol and calorimeter =(200s+10)5Rate of loss of heat =(200s+10)567cal/sRates of loss of heat in the two cases are equal∴ (200s+10)567=10 or s=0.62cal/g∘C