The specific heat of ice is 2.10kJ/kg degrees C. The heat of fusion for ice at 0

Question

The specific heat of ice is 2.10kJ/kg degrees C. The heat of fusion for ice at 0 degrees C is 333.7 kJ/kg. The specific heat of water is 4.186 kJ/kg degrees C, the heat of vaporization of water at 100 degrees C is 2256 kJ/kg, and the specific heat of steam is 2.020 kJ/kg degrees C. What is the final equilibrium temperature when 40.0 grams of ice at 0 degrees C is mixed with 8.00 grams of steam at 120 degrees C? Answer is 1.21 degrees C not sure how to get there.

Explanation / Answer

First we need to ensure that both ice and steam come to same state that is liquid. Heat absorbed by ice = 0.04*333.7 = 13.348 kJ Heat released by steam = 0.008*2.02*20 + 0.008*2256 = 18.3712 kJ Heat released > Heat absorbed hence the water at 0 degree will have a rise in temp 18.3712 - 13.348 = .04*4.186*t t = 30 degree Now we have 40 grams at 30 C and 8 grams at 100 C both in liquid state 0.04*4.186*(T - 30) = 0.008*4.186*(100 - T) => 5T - 180 = 100 - T => T = 43.33 C Please check you answer. I think its wrong. Even if we don't solve it we can just by reading the question tell that the final temperature would be around 40 C

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