Thermodynamics ice cube problem

[Jeffopolis]

First of all, hi, and thanks ahead of time... I've been working on this homework for almost a week, even asked the teacher for help but still haven't been able to figure out what I've been doing wrong/missing. its online work so ill copy paste it

Homework Statement

(a) Two 32 g ice cubes are dropped into 170 g of water in a thermally insulated container. If the water is initially at 27°C, and the ice comes directly from a freezer at -13°C, what is the final temperature (in Celsius) at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? The specific heat of water is 4186 J/kg·K. The specific heat of ice is 2220 J/kg·K. The latent heat of fusion is 333 kJ/kg.

Homework Equations

Q = cm(delta T)
Q = Lm

where c = specific heat, T = change in temp, m = mass of object, L = heat of fusion/vaporization

The Attempt at a Solution

I have a few questions that are similar but i think if i can solve one, then I'd be able to solve the others... I am not going to paste all my attempts because it'd take up 4 pages but ill summarize what i did

basically i know that:

0 = Q1 + Q2 + ... + Qn for all objects, since its in equilibrium/isolated location...

I'm assuming that there being 2 ice cubes, there will be 3 Q's

Q1 = water, cm(delta T) = 4186 * 170g * (Tfinal - 27)
Q2 = ice, cm(delta T) + heat of fusion?? [this part I am not sure if I am doing right] --- cm(delta T) + Lm = 2220 * 32 * (Tfinal +13) + 333 * 32
Q3 = same as Q2

adding em together and what not, i get:

Tfinal = (18279564 / 782660) = ~20.2ish... all my answers have been between like 20-20.5 for A and about 23 for B, neither of which have been right (there is a +/- 1 degree of error for part B, A must be exact)

Umm i have one more attempt to get the right answer on this one so yea... any help is greatly appreciated. It's due sunday and i have a gig tomorrow and sunday so whatever help i can get as soon as possible would be amazing

thanks again

 

[ehild]

Q2 = ice, cm(delta T) + heat of fusion?? [this part I am not sure if I am doing right] --- cm(delta T) + Lm = 2220 * 32 * (Tfinal +13) + 333 * 32

You can not jump from the initial temperature of the ice at the final temperature. The ice melts at 0°C, When melted, it is 0°C water, of mass equal to that of the ice, C the same as that of water.. So you calculate the heat needed to warm up the ice, add the heat of fusion, then add the heat to warm up the mass of water which was originally ice from 0°C to the final temperature.

Take care of the units. The specific heat capacities and the heat of fusion refer to 1 kg. You have to transform the mass values to kg-s first.ehild

 

[Jeffopolis]

so from what you said here's what I am getting:

C ice * M ice * (0 - [-13]) + L * M ice + C ice * M ice * (Tfinal - 0)

[heat to ice] + (heat of fusion) + [heat to melted ice] ?

and you do that with both cubes of ice?

edit: i just tried that and got the same answer (T0 = 0 duh lol)... can u help me with the equation please? that's where I'm having all the problem i think I am forgetting about something or not implementing some formula or something

edit2: ohhhh wait i was using the specific heat for the ice again on the second part. let me use the water's heat and see what happens.

edit3: bah nope was still wrong... the right answers were 0 for part A and 9 for part B... i still have no idea how those numbers were obtained tho... can't submit the correct work but if someone could show me how to get the right answer that'd be dope

 

[ehild]

It is not sure that the warm water can melt all the ice or even to warm it to the melting point. First calculate how much heat energy the warm water has above 0°C. Then find out how much heat is needed to warm up the ice to 0°C. If it is less than the heat "on store", you have 0°C ice and some heat. Next calculate how much heat is needed to melt the ice, and see if the remainder of the heat from the water is enough to that. If not, the final temperature is 0°C . If you have some heat left after all ice melted you have water to warm up.You will see that the warm water can melt one ice cube, but cannot melt two cubes.

ehild

 

[Jeffopolis]

If it is less than the heat "on store", you have 0°C ice and some heat.

I don't understand that part, how do you determine if it isn't enough?

using this example, Q for the ice to 0 degrees is 1847040J (for both ice cubes), and Q for the water is 19213740J from 27 to 0 degrees

How do you determine if that is enough heat or not enough heat? Or is my math wrong?

 

[ehild]

Your Q values are wrong. The specific heat capacities refer to 1 kg mass. Transform the mass values to kg.
The heat available from the warm water is somewhat more than the heat needed to warm up two cubes of ice. After the ice warmed up, there is some heat available to melt some. (How much?) How much heat is necessary to melt all the 0.064 kg ice? Take care of the units: The latent heat is given in kJ/kg.

ehild