# Help in understanding: Temperature, Heat, and Expansion

1. Dec 4, 2004

### eatubbs

Greetings,

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.

Eric

2. Dec 4, 2004

### savitri

eatubbs,

In problem no 2, you are missing the specific heat of water.
To find Q, the heat needed in Joules,
I would set it up this way:

Q = mass x sp.heat water x change in temp.

Last edited: Dec 4, 2004
3. Dec 5, 2004

### eatubbs

Savitri,

Thank you for the wonderful posting. In following your example, here is what I calcuated. My answer seems reasonable, and the process is much clear for this type of problem.

40 C = 313.15 K
20 C = 293.15 K
Difference in temperature = 20.00 K

So, I plug in 20 K into the following equation and get:

Water: 1.00 cal / (g*k) x 40g x 20K = 800 cal

Now, I convert calories to centigrade:

800 cal = 1.7636981
800 cal = 1.76 C

So, the change in temperature from 20 C to 40 C for 40g of water is 1.76 C

I am going to see if can solve problem number three in the same way, but I suspect there is an additional step or two. I will post my findings in a little while.

Thanks,

Eric

4. Dec 5, 2004

### arizonian

On problem # 2, set up a control volume, and calculate the energy in minus the energy out, which is zero. 20 degrees Centigrade is 20 degress centigrade, not 1.76.

For problem #3, you need to know the heat of fusion. When ice melts, it takes heat from the environment, which is the water. How much can you lower the temperature of liquid water before it is no longer liquid?

5. Dec 5, 2004

### eatubbs

Greetings,

I believe that I solved problem number 3, after much investigation:

80g of hot water at 90 degrees C in a large block of ice at 0 degrees C

Density of ice = 0.917 g/cm^3
Latent Heat of Fusion = 79.7 cal/g or 3.34x10^5 J/kg

For water cm change of T

1.00 cal/gC x 80g x 90C = 7200 cal

For ice Q = mL
solve for m = Q/L

m = (7200 cal) / (79.7 cal/g)
m = 90.338 g of ice will be melted

But, what how do I find out what the final temperature of the water in the cavity??

Thanks,

Eric

6. Dec 5, 2004

### eatubbs

Greetings again,

For problem 2

Would you mind explaining what you mean by a "control volume"? I believe that I misinterpreted the problem, which was quite dumb of me. Would I be correct in saying that it would take 800 calories of heat for the temperature change? The difference of temperature is indeed 20 degrees.. Where am I going wrong??

For Problem 3

Am I on the right track for problem 3? Or am I completely offtrack?

Thanks,

Eric

7. Dec 5, 2004

### savitri

eatubbs,

Stick with problem 2 first. The question asked is: How much HEAT is needed .. In other words we want to know Q=ENERGY=in Joules. This is not temperature. When you apply heat/energy, there is a change in temperature. ie from 20->40C.

using
Q = mass x sp.heat water x change in temp.

Fill in the blanks:

Q (in J) = __g x ___J/gC x ___C