A 0.5750-kg ice cube at -12.40 °C is placed inside a chamber of steam at 365.0 °

Question

A 0.5750-kg ice cube at -12.40 °C is placed inside a chamber of steam at 365.0 °C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 5.830 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. (Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.)

Explanation / Answer

heat gained by ice Q1 = M ice* ( Cice dT1 + Lf + Cw dT2) = 0.574*( 2100*12.4 + 335000+4183 t).....(1) Mw = 5.83*18 = 104.94Kg heat lost = Q2 = Mw *(Csteam 365+ Lv+ Cw (100-t) .........(2) heat loss = heat gain..........Q1 =Q2 substitute the values and find t

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