30gm of ice at -14 C is added to 200gm of water at 25 C find the equilibrium temperature. take spec. ht of ice = 0.5cal/g.C, spec. ht of water = 1cal/g.C and latent ht of water 80cal/g.

Kapil Khare
80 Points
5 years ago
Let us assume the final temperature to be T.
Now, we also know that T should be less than 25$\degree$C and greater than -14$\degree$C. Let us initially assume that our final state is water i.e. T is greater than 0$\degree$C. If after solving we get T less than 0$\degree$C then, we will say that our assumption is wrong and the final state is ice.
Magnitude of heat released by water to reach the equlibrium temperature = (200)(1)(25-T) calories
Magnitude of heat aborbed by the ice to reach 0$\degree$C = (30)(14)(0.5) calories = 210 calories
Magnitude of heat aborbed by the ice to meat to water = (30)(80) calories = 2400 calories
Magnitude of heat absorbed by this water to reach temperature T from 0$\degree$C = (30)(T-0)(1) calories
magnitude of total heat absorbed = (2610 + 30T) calories
magnitude of total heat absorbed = magnitude of total haet released
$\implies$                    2610 + 30T = 5000 – 200T
$\implies$                              230T = 2390
$\implies$                                    T = 10.39$\degree$C