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

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
 
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Arun (askIITians forum expert)

Dear Student,
Please find below the solution to your problem.

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