10 g of ice at -10 degree Celsius is added to 10 g of water at 85 degree Celsius. What is the final temperature and amount of ice left in the system?
Specific heat of water is = 4.2 J/(g.K)
Specific heat of ice is = 2.1 J/(g.K)
Latent heat of ice is = 336 J/g
Heat released by 10 g of water at 85 oC in converting to water at 0 oC is,
H1 = msθ = (10)(4.2)(85) = 3570J
Heat gained by 10 g of ice at -10 oC in converting into ice at 0 oC is,
H2 = (10)(2.1)(10) = 210 J
Suppose, ‘x’ g of ice now melts. So, heat gained by this ‘x’ g of ice in converting into water at 0 oC is,
H3 = xL = 336x J
Now, H3 + H2 = H1
=> 336x + 210 = 3570
=> x = 10 g
Mass of ice remaining is = 10 – 10 = 0 , no ice will remain in the system.
Thus, in the resulting mixture the ice remaining is 0 g and the temperature of the mixture is 0 oC.