# 10 grams of ice at zero degree celsius is mixed with 100 grams of water at 50 degree Celsius what is the resultant temperature of the mixture

Arun
25757 Points
5 years ago
Latent heat of fusion of ice = 336 J/g
specific heat of water = 4.186 J/g degreeC
First 10g ice melts and then its temperature increases by gaining heat from 100g water. let the resultant temperature is T.
Energy gained by 10g ice = energy lost by 10g water
10*(336) + 10*4.186*(T-0) = 100*4.186*(50-T)
3360 + 41.86 T = 20930 - 418.6T
(41.86+418.6)T = 20930 - 3360
460.46T = 17570
T = 38.15 degree celsius

Regards
Rishi Sharma
3 years ago
Dear Student,

Latent heat of fusion of ice = 336 J/g
specific heat of water = 4.186 J/g degreeC
First 10g ice melts and then its temperature increases by gaining heat from 100g water.
let the resultant temperature is T.
Energy gained by 10g ice = energy lost by 10g water 10*(336) + 10*4.186*(T-0)
= 100*4.186*(50-T) 3360 + 41.86 T
= 20930 - 418.6T (41.86+418.6)T
= 20930 - 3360 460.46T
= 17570 T
= 38.15 degree celsius

Thanks and Regards