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This is my first post and I hope that they way I present my questions will be acceptable to all who might read them. I will present as much information as possible because I want to understand the concepts that do not make much sense to me.

I had a total of three problems that I couldn't work out, but I was able to solve one of them, and I am quite happy, but these other two I cannot seem to get solved and/or figure out.

First Question: The one that I solved:

How much heat is required to raise the temperature of 40g of iron from 20 degrees C to 40 degrees C? (I was given the specific heat of iron is 0.12 cal/g C)

This is how I set the problem up:

heat gained by water = heat lost by iron

(cm change of T)water = (cm change of T)iron

(1)(40g)(T-20) = (0.12 cal/g C)(40g)(40 - T)

40(T-20) = 4.8(40 - T)

40T - 800 = 192 - 4.8T

40T + 4.8T - 800 +800 = 192 + 800 -4.8T +4.8T

44.8T = 992

T = 992/44.8

T = approx 4.285

T = 4.29 degrees C of heat gained

This problem took me quite a while, but I was happy when I figured it out.

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Now, comes the second problem that I cannot figure out, but I have worked the parts of the problem that I know how to do.

How much heat is required to raise the temperature of 40g of water from 20 degrees C to 40 degrees C?

This is how I set the problem up:

heat gained by water = heat lost by water

(cm change of T)water = (cm change of T)water

(1)(40g)(T-20) = (what goes here?)(40g)(40 - T)

40(T-20) = (what goes here?)(40)(40 - T)

40T - 800 = I cant fill this in yet because of the unknowns

T =

T =

T =

Mg first attempt at this problem was to put a "1" where the unknown was, but that didn't work, eg.

(1)(40g)(T-20) = (1)(40g)(40 - T)

40(T-20) = 40(40 - T)

40T - 800 = 1600 - 40T

From here, this problem is not solvable because when I combine the T's together, the resulting number is 0, thus -800 is not equal to 1600. Any suggestions?

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My last problem really stumped me, because I do not know how to set it up correctly.

IF 80 grams of hot water at 90 degrees C is poured into a cavity in a very large block of ice that is at 0 degrees C, what will be the final temperature of the water in the cavity? How much ice must melt to cool the hot water down to this temperature?

heat gained by water = heat lost by water in ice

(cm change of T)water = (cm change of T)water in ice

(1)(80g)(T-90) = (something goes here)(90g)(something - T)

80(T-90) = (something goes here)(90)(something - T)

80T - 7200 = I cant fill this in yet because of the unknowns

T =

T =

T =

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I hope that I have presented my questions in a suitable and respectable way that will not offend anyone who reads them. I want to understand the concepts more than just the answers, so please, if you will help me set up the problems, I will gladly solve them to the best of my ability.

Many thanks in advance,

Eric

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# Homework Help: Help in understanding: Temperature, Heat, and Expansion

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