Question

The specific heat of water in its solid phase (ice) is 2090 J/(kg K), while in the liquid phase (water) its specific heat is 4190 J/(kg K). Water's latent heat of fusion is 333,000 J/kg.

If you have a 2kg block of ice at -90 degrees C and you add 1,000,000 J of heat, what is its new temperature?

Answer 1

Solution :

Given :

mass (m) = 2 kg

T_i = - 90^oC

C_ice = 2090 J/kg K

C_water = 4190 J/kg K

L_fusion = 333,000 J/kg

and, Q = 1,000,000 J

.

Since, Q = m C_ice (90) + m L_fusion + m C_water (T_f)

1,000,000 J = (2 kg)(2090 J/kg K)(90) + (2 kg)(333,000 J/kg) + (2 kg)(4190 J/kg K) T_f

1,000,000 J = (376200 J) + (666000 J) + (8380) T_f

(8380)T_f = - 42200 J

Which means that the given  1,000,000 J of heat can not melt all the ice.

Thus the final temperature will be : 0 ^oC