To cool a very hot piece of 4.0-kg steel at 900*C, the steel is put into a 5.0-k

Question

To cool a very hot piece of 4.0-kg steel at 900*C, the steel is put into a 5.0-kg water bath at 20*C. What is the final temperature of the steel-water mixture?

The book says that the specific heat (c) for water is 4186 J/kg x *C and steel is 460 J/kg x *C.

I just found out that the answer is 195*C, but I still don't know how to get that. Since it is a phase change question maybe you're suppose to use the latent heat of vaporization (22.6 x 10^5 J/kg) for water in the problem.

The closest I've ever gotten to the answer is 192*C.

Explanation / Answer

First we need to determine if the heat energy required to lower steel temperature from 900oC to 100oC is greater than the heat required to raise the water temperature 20oC to 100oC

Qs = mscs(Tf-Ti)  = (4kg)(460 J/kg*C)(900oC-100oC) = 1472000J = 1.472*106J

Qw = mwcw(Tf-Ti)  = (5kg)(4186 J/kg*C)(100oC-20oC) = 1674400J = 1.6744*106J

This shows us that it takes more energy to raise the water temperature to boiling point, than to lower the steel temperature to 100oC (boiling point of water). So the final temperature of the steel-water mixture would be lower than the boiling point of water 100oC. We dont have to use the latent heat of vaporization for water for this problem.

 

Qs + Qw = 0

Qs = - Qw

msCs(Ts) = -mwCw(Tw)

(4kg)(460 J/kg*C)(Tf - 900oC) = -(5kg)(4186 J/kg*C)(Tf-20oC)

 1840 Tf -1656000 = -20930 Tf + 418600

2074600 = 22770 Tf

Tf = 91.1oC

 

I dont know who told you answe is 195oC, I am very sure this is how you do this problem. 

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