Question

Find the result of mixing 10g of ice at -10 degree C with 10 g of water at 10 degree C. Specific heat cap. of ice= 2.1 J/g K. Specific latent heat of ice= 336 J/g. Specific heat cap. of water=4.2 J/g K

Answer 1

here heat will flow from water to ice.

heat lost by water , H = ms∆T

= 10g × 4.2J/g/K × (10 - 0)

= 420J

heat gained by ice , H' = ms'∆T

= 10g × 2.1J/g/K × {0 - (-10)}

= 210 J

H > H'

hence, some amount of ice is melting.

Let m' is mass of ice melted.

H = H' + m'L

420 J = 210J + m' × 336 J/g

or, 210 = 336 × m'

or, m' = 210/336 = 0.625g

hence, amount of water = 10 + 0.625 = 10.625g

amount of ice = 10 - 0.625 = 9.325g

Answer 2

Please take reference of the attachments