Solving Heat Transfer Puzzle: Cooling Steel from 452°C to 100°C

[Yomna]

Hey People...um this is a question i got for school...i know what was used to solve it but i can't understand how that method was used and why they did that...

it says:
A foundry operator finds that it takes 55.3MJ of heat to heat a 286 kg mass of an alloy steel from 22 degrees C to 452 degrees C.

The first thing they ask us is to find the specific heat capacity for the steel...which is pretty straight forward..:
c= Q/m*delta T
Q= 55.3 MJ
m=286
delta T=430
(55.3*10^6)/(430*286)
=4.5*10^2 J per Kg per kelvin..

The next bit is what's bothering me...
it says:

If the foundry worker cools the steel by pouring 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 C, would cool the hot steel down to 100 degrees C?

I really don't know what to do here...
Can someone please help? and like..explain?
Thank u so much.
Yomna

 

[dextercioby]

HINT:What is the heat that the steel gives away to drop its temp from 452°C to 100°C...?

Daniel.

 

[Yomna]

Well..i thought of that.
I use thermal eguilibrum ryt?
In the answers it is written:
Q(total)=-55.3x10^6=mc delta T(of water) + mL (water)
I just don't understand why they did that. When i tried this i just did:

Q(total)=mc delta T (of water)...can u tell me why they added the heat taken to evaporate the water?
And why did they put -55.3 instead of 55.3?

Thanks.
Yomna

 

[dextercioby]

I don't know what wicked sign conventions they have,but here's how i do it.

The heat the water receives to reach boiling point is taken from the cooling metal.The heat gave away by the metal while cooling is simply

$$Q_{gave \ away}=:Q=m_{metal}c_{metal}(452 \deg \ C-100 \deg \ C) \ [J]$$

U get a #.Okay u know that this heat is neceassary for an unknown mass of water to reach boiling point when heated from 22°C...So

$$Q=m_{water}c_{water}(100 \deg \ C-22 \deg \ C) \ [J]$$

U know Q (found above),u can take $c_{water}\approx 4185J \ \left(Kg\right)^{-1} (deg \ C)^{-1}$ and u need to find $m_{water}$.

So do it.

Daniel.

 

[Yomna]

Last one :D

Hey..sorry ...but just one last question..
does latent heat have anything to do with this question then?
Thanks
Yomna

 

[marlon]

Hey..sorry ...but just one last question..
does latent heat have anything to do with this question then?
Thanks
Yomna

You can answer this question for yourself if you just realize that latent heat is the heat necessary to make the actual phase-transition. For example when you heat up ice (-5°) to water (+5°) you will need to look at three things :

1) heat necessary for the ice to go from -5° to 0°
2) heat necessary for ice to transform into water (ie latent heat)
3) heat necessary for water to go from 0° to 5°

marlon

 

[dextercioby]

"(...)the water will heat up to its boiling point, then it will boil.".So it mentions boiling,too.My analysis,didn't include it.U may take it into account,if u want to.In that case,the total mass of water (determined above) would be different.

Daniel.

 

[Yomna]

Well...i understand it now
Cos it asked for the minimum mass of water needed...
So they added the energy taken to heat it to 100 degrees then the energy taken to evaporate that minimum amount of water...
so that's why they did...

Q= mc delta T + mLv


So i think that was just my main concern..i didnt know why they added the latent heat too.

Thanks for all ur help :D

Yomna.

 

[dextercioby]

Because adding that heat reduce the amount of water.If u did compute the water needed in the first case and now in the second,u'd have seen the difference & the relevance of the formulation "minimum amount of water".

Daniel.