Rishi Sharma
Last Activity: 4 Years ago
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