Water vapor with mass, m v = 5g, and temperature, T v = 120 C, is bubbled into t

Question

Water vapor with mass, mv = 5g, and temperature, Tv = 120 C, is bubbled into the bottom of a container of water with mass, mw = 100g , and temperature, Tw = 20 C . No water vapor leaves the top of the water.

What is the final temperature, T, of the water?

Note: Some properties of water that you might need are to one significant figure: cice = 2 kJ/kg/K, cwater = 4 kJ/kg/K, cvapor = 2 kJ/kg/K, Lfusion = 300 kJ/kg, and Lvapor = 2000 kJ/kg. To make your calculations easier, your answer need only be accurate to one significant figure. (You can neglect any increase in vapor pressure above the water.)

Explanation / Answer

heat lost by vapor = heat gained by water

mv*cvapor*dt1 + mv*Lv + mv*cwater*dt2 = mw*cw*dt3

(0.005*2000*(120-100)) + (0.005*300000) + (0.005*4000*t)

= (0.1*4000*(t-20))

final temperature = 25.53 degrees

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