1.

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» OK ok ok ok ok ok ok ok ok ok ok


Discussion

No Comment Found

Related InterviewSolutions