Question

80 gm of ice at 10°C is mixed with 460 gm of liquid water at 20 °C and 10 gm steam at
100°C. The container is insulated so that no heat is lost. Find the equilibrium temperature
of the contents (in °C).
Given: Speci c heat of ice = 0.5 cal/gm°C ; Speci c heat of water 1 cal/gm°C
Latent of fusion(water-ice): 80 cal/gm ;
Latent heat of evaporation(water-steam): 540 cal/gm

Answer 1

Answer:

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