Mass of vapor per hour for heating water

[songoku]

Homework Statement

To heat 10 kg water per hour from 20 degree Celsius to 80 degree Celsius, a 150 degree Celsius vapor from a kettle is passed through a pipe which is put into water. The vapor will condense and brought back to kettle as water at 90 degree Celsius. How much vapor is needed per hour?

Relevant Equations

Black principle

Q = m.c.ΔT

Q = m.Lv

I thought about using Black principle to solve this question but I am confused about the final temperature of system

Q released by vapor = Q absorbed by water

m_vapor . c_vapor . ΔT_vapor + m_vapor.Lv + m_vapor . c_water . ΔT₂ = m_water . c_water . ΔT_water

But what is the final temperature of the system? The water is heated from to 80^oC and vapor cools down to 90^oC so final temperature of water and vapor is not the same?

So this means that Q released by vapor = Q absorbed by water does not imply that the final temperature of the vapor and water must be the same?

Thanks

 

[haruspex]

Q released by vapor = Q absorbed by water does not imply that the final temperature of the vapor and water must be the same?

Quite so.

 

[songoku]

Thank you very much haruspex