Answer:
4.88 Cals per degree celsius
Explanation:
We have taken heat of fusion of ice = 80 cals / g
We have taken speciic heat of water = 1 cal/g per degree celsius
In this experiment , let the heat capacity of calorimeter be X.
Heat gained by ice
heat gained in melting + heat gained in getting warmed
= mass x latent heat + mass x specific heat x rise in temperature
= 17.69 x 80 + 17.69 x 1 x ( 12.9 - 0 )
= 1643.4 Cals
Heat lost by water
= mass x specific heat x fall in temperature
98.67 x 1 x ( 28.77 - 12.9 )
= 1565.89 Cals
Heat lost by calorimeter
heat capacity x fall in temperature
X x ( 28.77 - 12.9 )
Heat gained = heat lost
1643.4 = 1565.89 +15.87X
X = 4.88 Cals per degree celsius