Question

Find the result of mixing 10g of ice at -10°C with 10g of water at 10°C.Specific heat capacity of ice=2.1 J/gK ,specific latent heat of ice =336J/g and specific heat capacity of water=4.2J/gK

Answer 1

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 10°C in converting to water at 0°C is,

H1 = msθ = (10)(4.2)(10) = 420 J

Heat gained by 10 g of ice at -10°C in converting into ice at 0°C 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 = x336 = 336x J

Now, H3 + H2 = H1

=> 336x + 210 = 420

=> x = 0.625 g

Mass of ice remaining is = 10 – 0.625 = 9.375 g

Thus, in the resulting mixture the ice remaining is 9.375 g and the temperature of the mixture is 0°C.