[oceanmd]

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]

yes this is correct. No phase change is involved, thus you only use 2.03 as the specific heat of ice.

[oceanmd]

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]

you know i'm honestly not sure. let me refer to a book to check this one

[enahs]

Water and Ice can and do coexist at 0^oC.

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

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

[flightman233]

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]

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.

Add the two steps together and you will have your answer.

[oceanmd]

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]

oceanmd: you are mixing kJ and J, but the general idea is OK.

[oceanmd]

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]

q3 = 11014.8 J = 11014.8 x 0.2390 = 2633 cal

Huh?

[oceanmd]

The teacher asked for the answer in cal. Is it wrong?

[ARGOS++]

Dear Oceanmed;

Use it as your “Guardrail”:   "Heat Capacity”


Good Luck!
                    ARGOS⁺⁺

[Borek]

The teacher asked for the answer in cal. Is it wrong?

No, that's OK, you have just nor signalled what you are doing and why and your q3 looked completely random.