Heat: word problem?

  • Thread starter Dx
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  • #1
I missed the test question can anyone explain what they did to solve for this.

In grinding a steel knife blade(SH=.11 cal/g*C) the metal can get as hot as 400C. If the blade has a mass of 80g, what is the minimum amount of water needed at 20C if the water is not to rise above the boiling point when the hot blade is quenched in it?

answers were: 22, 33, 44 or 55g

I tried using Q=mc[del]T then the difference of the two but didnt work that way, can anyone please help me?
TY
Dx :wink:
 

Answers and Replies

  • #2
quantumdude
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Originally posted by Dx
I tried using Q=mc[del]T then the difference of the two but didnt work that way, can anyone please help me?

I don't know what you mean by "the difference of the two", but to do this problem at the most basic level you have to assume that all the heat lost by the steel is gained by the water.

Heat lost by steel: qS=mScSΔTS
Heat gained by water: qW=mWcwΔTW

You have the initial temps of both the steel and the water. Assume that the steel and water come to thermal equilibrium. The only unknown is mW.
 
  • #3
HallsofIvy
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You know the temperature of the blade is 400 degrees C, the mass is 80 grams and the specific heat is .11 cal/g*C so the total heat in the blade is 400*80*.11= 3520. How many calories does it take to raise 1 gram of water 1 degree? (That's what Tom meant by sw, the specific heat of water. Good job, Tom.) That's physical constant that should be given in your book just as the specific heat of steel was.

Once you know that, how many calories are required to raise the temperature from 20 to 100 degrees?

Finally, once you know that, how many grams of water do you need to account for all those calories?
 

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