HEAT GAINED = HEAT LOST
use eqn mcdt....m is the mass [0.01 kg], c is the specific heat capacity of water [4200 J kg-1 K-1] and dt is change in temp.
T1 = -10 0C = 263K and T2 = 85 0C = 358K
Let T be the final temp.
Then, 0.01 kg* 4200 * (T-263) = 0.01 kg * 4200 * (358-T) [Heat gained = heat lost]
Solving the eqn gives
T-263 = 358-T
2T = 358+263 = 621
or, T = 310.5K = 37.35 0C
FINAL TEMP, T = 310.5K = 37.35 0C
NO ICE WILL LEFT IN THE SYSTEM AS IT COMPLETELY BECOMES WATER
