Changes of state and latent heat

[shannon.leigh]

Homework Statement

A foundry operator finds that it takes 55.6 MJ of heat to heat a 286 kg mass of an alloy steel from 22 degrees C to 452 degrees C. Specific heat capacity of the steel is 452J/kg/K.
If the foundry worker cools the steel by poring water onto it, the water will heat up to its boiling point, then it will boil. What minimum mass of water, initially at 22 degrees celsius, would cool the hot steel down to 100 degrees celsius?

Homework Equations

I figured that
delta Q= delta T*c*m
and
delta Q=L(vaporisation)*m
would probably be relevant

The Attempt at a Solution

I went. . .
delta Q(of steel) =delta T*m*c
=352*452*286
= 45503744 J therefore Delta Q(of water)= 45503744 J

Delta Q(of water)=delta T*m*c
m(of water)=Delta Q/(delta T*c)
=45503744/(78*420)
=139 kg

But the book said the mass of water was 21.5 kg. :S
I think I need to incorporate the Latent heat of vaporistion of water into this somewhere, but I am way to confused to work out how. . .
Any help would be soooo apreciated!

Thankyou!

Shannon

 

[myuugen]

The Q leaving the steel must equal the Q entering the water. The water heats up, then changes phase. The heat necessary to change a certain mass of water from liquid to steam is your latent heat. Therefore, you're missing something in your delta q of water

 

[shannon.leigh]

thanks! big help :)