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### Topic: Need help on heat absorbed  (Read 7489 times)

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#### oceanmd

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##### Need help on heat absorbed
« on: March 30, 2008, 10:27:29 PM »
Please check the solution. Thank you very much.

32 g of water is heated from ice at -5 degrees C to water at 0 degrees. How much heat is absorbed?

My solution: 2.03 (c ice) x 32 x 5 = 324.8 J

Is this it? Or do I need to do something else? Thank you

#### Kyle1990

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##### Re: Need help on heat absorbed
« Reply #1 on: March 31, 2008, 12:23:14 AM »
yes this is correct. No phase change is involved, thus you only use 2.03 as the specific heat of ice.
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#### oceanmd

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##### Re: Need help on heat absorbed
« Reply #2 on: March 31, 2008, 12:33:47 AM »
So I do not need to do q2 = 6.01 x 32 = 192.32
q total = 517.12 J

At 0 degrees C ice starts to melt. Are you sure there is no change of state?

Thank you again for your help.

#### Kyle1990

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##### Re: Need help on heat absorbed
« Reply #3 on: March 31, 2008, 03:18:04 PM »
you know i'm honestly not sure. let me refer to a book to check this one
"Theories are nets cast to catch what we call 'the world': to rationalize, to explain, and to master it. We endeavor to make the mesh ever finer and finer."
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#### enahs

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##### Re: Need help on heat absorbed
« Reply #4 on: March 31, 2008, 08:24:09 PM »
Water and Ice can and do coexist at 0oC.

So if it says Ice at 0oC you do not have to account for the enthalpy of fusion. If it says water at 0oC you do.

Do you have ice or water are 0oC in your question?

#### flightman233

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##### Re: Need help on heat absorbed
« Reply #5 on: April 01, 2008, 11:23:05 PM »
Enhas is pointing you in the right direction.

Take a closer look at your problem at hand.

You have two things going on here, 1 ice at -5 deg C going to 0 deg C
and ice at 0 deg C turning completely to water at 0 deg C.

Two equations are required to solve the problem.

Q1 = m(ice)C(ice)
Q2 = m(water)C(water)(del T)

Q1 + Q2 = heat absorbed by the system.

#### flightman233

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##### Re: Need help on heat absorbed
« Reply #6 on: April 02, 2008, 07:17:43 AM »
I must make an ammendment to what I stated previously

Q2 = m(ice)C(ice)(del T)

I just realized that I put water, but we aren't raising the temperature of water in any way.

#### oceanmd

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##### Re: Need help on heat absorbed
« Reply #7 on: April 13, 2008, 01:39:30 PM »
I still do not have the clear understanding how to solve this kind of problem:
32 g of water is heated from ice at -5 degrees C to water at 0 degrees. How much heat is absorbed?

We need to find q1 and q2 and add them together.
1. q1 = c(ice) x m (mass) x delta T ; q1 = 2.03 x 32 x 5 = 324.8 J
2. As it is water at 0 degrees, we need to use H fus, which is 6.01 KJ/m. This is the part I think I made the mistake before. As it is 6.01 KJ per mole, I need to convert 32 g of H2O into moles,32 / (2+16) = 1.78 mol
q2 = 6.01 x 1.78 = 10.69 J
q total = 324.8 + 10.69 = 335.49 J
Please check and explain if I am doing it wrong. Thank you.

#### Borek

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##### Re: Need help on heat absorbed
« Reply #8 on: April 13, 2008, 02:16:58 PM »
oceanmd: you are mixing kJ and J, but the general idea is OK.
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#### oceanmd

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##### Re: Need help on heat absorbed
« Reply #9 on: April 13, 2008, 02:48:50 PM »
Borek,

How does this look now?
q1 = 324.8 J
q2 = 6.01 x 1.78 = 10. 69 KJ = 10690 J
q3 = 11014.8 J = 11014.8 x 0.2390 = 2633 cal

Thank you very much.

#### Borek

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##### Re: Need help on heat absorbed
« Reply #10 on: April 13, 2008, 03:04:17 PM »
q3 = 11014.8 J = 11014.8 x 0.2390 = 2633 cal

Huh?
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#### oceanmd

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##### Re: Need help on heat absorbed
« Reply #11 on: April 13, 2008, 03:15:50 PM »

#### ARGOS++

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##### Re: Need help on heat absorbed
« Reply #12 on: April 13, 2008, 03:18:14 PM »
Dear Oceanmed;

Use it as your “Guardrail”:   "Heat Capacity

Good Luck!
ARGOS++

#### Borek

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##### Re: Need help on heat absorbed
« Reply #13 on: April 13, 2008, 03:52:16 PM »