Finding mass of ice in latent heat of fusion problem

In summary: I think I'm just going mad with all this maths. I'm not usually this bad but I have a lot on my plate at the moment and I'm totally stressed. I think I'll just have to hand this in tomorrow and see what happens. I've done my best. thanks for your help anyway!I'm sorry, but I cannot provide a summary for this conversation as it does not contain any relevant information.
  • #1
aussie-girl
5
0

Homework Statement



The question is: a 6 pack of cola cans contains m grams of liquid. you want to put ice into a perfectly insulated box to cool it down from Ts to Tf. The ice initially has temp Ti. So (a)how much ice would you need as a minimum. (ignore presence of cans and final state of ice will be water). (b) calculate the mass of ice initially at -5 deg celcius needed to cool 2.1kg of canned cola from 23 deg celcius to 4 deg celcius.

Homework Equations



specific heat of water = 4.18J
Lf of water = 333.55Jkg-1
mass(cola) x C (water) (Tf - Ts) = mass(ice) x C(ice) (0 - Ti) + mass(ice) x Lf + mass(ice) x C(water) (Tf - 0)


The Attempt at a Solution



I've worked out that the amount of energy needed to melt 'm'kg of ice = m(ice) x Lf(water) or Q = mL. also that heat lost by cola = heat gained by ice

I realize there are 3 stages involved:
(1) heat needed to raise ice to O deg celcius
(2) heat needed to melt ice
(3) heat needed to raise temp of water (from melted ice) to Tf from O deg celcius.

Then I get lost. How do I find the mass of the ice to begin with? Do I rearrange the equation? If so, I have no idea how to start rearranging a long complex equation like this.

Thank you for any guidance on this.
 
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  • #2
Welcome to PF!

Hi aussie-girl ! Welcome to PF! :smile:
aussie-girl said:
a 6 pack of cola cans contains m grams of liquid. you want to put ice into a perfectly insulated box to cool it down from Ts to Tf. The ice initially has temp Ti. So (a)how much ice would you need as a minimum. (ignore presence of cans and final state of ice will be water). (b) calculate the mass of ice initially at -5 deg celcius needed to cool 2.1kg of canned cola from 23 deg celcius to 4 deg celcius.

I realize there are 3 stages involved:
(1) heat needed to raise ice to O deg celcius
(2) heat needed to melt ice
(3) heat needed to raise temp of water (from melted ice) to Tf from O deg celcius.

Then I get lost. How do I find the mass of the ice to begin with? Do I rearrange the equation? If so, I have no idea how to start rearranging a long complex equation like this.

Thank you for any guidance on this.

call the mass of ice M, and write out the equations …

then we'll see how to rearrange them :wink:
 
  • #3


Hi tiny-tim, nice to meet you!

So done as suggested and here is the result:

part (a):

m(cola) x C(water) (Tf - Ts) = M(ice) x C(ice) (0 - Ti) + M(ice) x Lf + M(ice) x C(water) (Tf - 0)

part (b):

2.1kg x 4.186 x (4 - 23) = M(ice) x 2.093 x (0 - -5) + M(ice) x 333.55 + 4.186 (4 - 0)


I think I've substituted in the right figures. Please advise how I now proceed?

Thanks, appreciate your time and help!
 
  • #4
aussie-girl said:
2.1kg x 4.186 x (4 - 23) = M(ice) x 2.093 x (0 - -5) + M(ice) x 333.55 + 4.186 (4 - 0)

Hi aussie-girl! :smile:

(i haven't checked your figures, but …)

put M(ice) outside a bracket on the right-hand side …

M(ice) x [2.093 x (0 - -5) + 333.55 + 4.186 (4 - 0) ] :wink:
 
  • #5
Hi again

I'm confused by what you said. I can see that you've taken the second M(ice) out of the equation and assume you mean to put that outside a bracket on the right-hand side? but that doesn't make sense to me. are you showing me what you mean by this:

M(ice) x [2.093 x (0 - -5) + 333.55 + 4.186 (4 - 0) ] but that can't be right because the second M(ice) is missing.

I've been thinking about this and I can't figure it out. maybe I'm just being thick but it is 3.30am over here! can you please explain more? I'm not expecting you to solve it for me but I need more explanation.

Thanks again...
 
  • #6
aussie-girl said:
I've been thinking about this and I can't figure it out. maybe I'm just being thick but it is 3.30am over here!

ah yes … i should have noticed that …

the clue's in the name, isn't it? :biggrin:

Ma + Mb = M(a + b) …

(oh, and I should have written M(ice) x [2.093 x (0 - -5) + 333.55] + 4.186 (4 - 0) :redface:)

that may not mean anything to you right now, but …

get some sleep! :zzz:

… and it'll all be clear in the morning. :smile:
 
  • #7
Hi again

Just before I go to bed (I need to know this stuff for a paper that's due tomorrow!) can I see if this is what you mean:

2.1kg x 4.186 x (4 - 23) = M(ice) x [2.093 x (0 - -5) + 333.55] + 4.186 (4 - 0)

-167.021 = M(ice) x 362.2

so to get M(ice) on its own via rearrangement, you get:

M(ice) = -167.021 / 362.2 = -0.46

I think I'm going crazy :'( this doesn't seem right! the answer is a negative. God I don't know! I think I'll go to bed now...

thanks again tiny-tim.
 

1. How do I calculate the mass of ice in a latent heat of fusion problem?

To calculate the mass of ice in a latent heat of fusion problem, you will need to use the formula: mass of ice = (latent heat of fusion * amount of energy added) / heat of fusion. The latent heat of fusion is a constant value for a specific substance, while the heat of fusion can be found in a reference table for different substances.

2. What is the unit of measurement for the mass of ice in a latent heat of fusion problem?

The unit of measurement for mass of ice in a latent heat of fusion problem is typically in grams (g) or kilograms (kg). This will depend on the units used for the latent heat of fusion and heat of fusion values in the calculation.

3. Can the mass of ice in a latent heat of fusion problem be negative?

No, the mass of ice in a latent heat of fusion problem cannot be negative. This value represents the amount of ice present, and a negative mass would not make sense in this context.

4. How do I determine the latent heat of fusion for a specific substance?

The latent heat of fusion for a specific substance can be found in a reference table or by conducting experiments. This value is a constant for a specific substance and can vary for different substances.

5. What other factors should I consider when calculating the mass of ice in a latent heat of fusion problem?

When calculating the mass of ice in a latent heat of fusion problem, you should also consider the initial temperature of the ice and the final temperature of the melted ice. These values will affect the amount of energy needed to melt the ice and, consequently, the mass of ice present.

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