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shannon.leigh
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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 celcius, would cool the hot steel down to 100 degrees celcius?
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