Find resulting temperature, heat quantity problem

[aelarz]

Homework Statement

Ten kilograms of steam at 100 degrees C is condensed by passing it into 500 kg water at 40 degrees C. What is the resulting temperature?

Homework Equations

sum of all heats must be = zero
heat of vaporization for water = 540 cal/g
change Q = c*m* change in T where Q is heat, c is specific heat, m is mass, and T temp.

The Attempt at a Solution

convert kg to g, .01 g steam and .5 g water

heat to convert steam to H20 + heat to cool steam + heat to warm water = 0
.01(540) + .01(T - 100) +.5 (T - 40) = 0
5.4 +.01T -1 +.5T - 20 =0
T = 30.6 C
or
heat to convert stem to water+heat to cool steam to 40C+heat to warm all h20=0
5.4 + .01(-60) +.51 (T-40) = 0
T = 30.6 C
... but the answer is 51.8 C
How is this so?

Thanks to any help!

 

[ideasrule]

The Attempt at a Solution

convert kg to g, .01 g steam and .5 g water

heat to convert steam to H20 + heat to cool steam + heat to warm water = 0
.01(540) + .01(T - 100) +.5 (T - 40) = 0
5.4 +.01T -1 +.5T - 20 =0

There are several problems with this. First, 1 kg=1000 g, so 10 kg of steam is 10,000 g. Second, the equation for heat transferred is mc*delta_T. You forgot the "c". Third, you have to be very careful with the sign of your terms. Every term should represent a difference in heat between the final and initial states. So your first term should be negative because the steam lost heat, the second should be negative as well (which it is) because the condensed water also looses heat, and the third should be positive because the cold water gains heat.

 

[aelarz]

Thank you... I did not write in c because c for h20 is 1... so...
heat to convert steam to h20+heat to cool steam+ heat to warm h20 = 0

-10,000(540)+10,000(1)(T-100)+500,000(1)(T-40) = 0
-5,400,000 + 10,000T -1,000,000 +500,000T-20,000,000 = 0
26,400,000 =510,000T
T=51.8 C

but which third term were you referring to that needs to be positive?