How much ice to cool down this water?

[Polarbear10]

Homework Statement

Ice --> water

Relevant Equations

Q=c*m*∆T
Q=m*Heat of fusion

Hello!
I'm having some problems solving a task..
"Ice with temperature -12°C is used to cool down water from 98°C. to 55°C. . Specific heat capacity for ice = 2097J/kgK, Secific heat capacity water: 4180 j/kgK. Heat of fusion for water is 334,4kJ7K. The mass of the water is 20kg."
∆T water = 55°C - 98°C = -43°C
∆T ice = -55°C - (-12°C) = 67°C
How much ice do we need to cool down the water?
The correct answer is supposed to be 6,1 kg of ice.
My thoughts: Heat gained + heat loss = 0 --> where ice melting (Q1) + icewater warming (Q2) (being the heat gained) and the heat loss being water cooling (Q3)
So I made these equations:
Q1: Heat of fusion water * mass of ice = 334400 * x = 334400x
Q2: Specific heat capacity for ice* m * ∆T = 2097 * 67 * x = 140499x
Q3: 20 * 4180 * -43 = -3594800J
Q1+Q2+Q3 = 0
334400x + 140499x - 3594800 = 0

474899x = 3594800
/474899 = /474899
x (mass of ice) = 7,57 kg
So I can't seem to get the correct answer.. is my ∆T for ice wrong?
Appreciate any help! :)

 

[DrClaude]

When the ice melts, it becomes water, so it no longer has the same heat capacity.

 

[Polarbear10]

When the ice melts, it becomes water, so it no longer has the same heat capacity.

I tried using heat capaity for water, ended up with the mass of ice being 5,8kg..

 

[DrClaude]

I tried using heat capaity for water, ended up with the mass of ice being 5,8kg..

You were not using the wrong number, you were using the wrong approach. Since ice water and liquid water do not have the same heat capacity, you need to divide the process in more steps.

 

[Polarbear10]

You were not using the wrong number, you were using the wrong approach. Since ice water and liquid water do not have the same heat capacity, you need to divide the process in more steps.

Any tips on how I do it?

 

[PeroK]

Any tips on how I do it?

The ice gets to freezing point, the ice melts, and then the waters mingle.

 

[PeroK]

How much ice do we need to cool down the water?
The correct answer is supposed to be 6,1 kg of ice.

That's what I get. I took an algebraic approach and put the numbers in a spreadsheet. I didn't do any calculations myself.

 

[PeroK]

I took $Q_w$ to be the heat lost by the water:$$Q_w = M\Delta T_3 c_w$$where $M = 20 \ kg$, $\Delta T_3 = 43 \ K$ and $c_w$ is the specific heat of water.

Then, I took the mass of ice to be $m$ and calculated the three stages of heating:
$$Q = Q_1 + Q_2 + Q_3 = mq_1 + mq_2 + mq_3$$Then I equated the two and took out the common factor of $m$, so I ended up with:$$m = \frac{Q_w}{q_1 + q_2 + q_3}$$I find all those numbers get confusing. I prefer to let a computer do the arithmetic.

You need to find the correct expressions for the three stages of what started as ice being heated.

 

[Philip Koeck]

Homework Statement:: Ice --> water
Relevant Equations:: Q=c*m*∆T
Q=m*Heat of fusion

Hello!
I'm having some problems solving a task..
"Ice with temperature -12°C is used to cool down water from 98°C. to 55°C. . Specific heat capacity for ice = 2097J/kgK, Secific heat capacity water: 4180 j/kgK. Heat of fusion for water is 334,4kJ7K. The mass of the water is 20kg."
∆T water = 55°C - 98°C = -43°C
∆T ice = -55°C - (-12°C) = 67°C
How much ice do we need to cool down the water?
The correct answer is supposed to be 6,1 kg of ice.
My thoughts: Heat gained + heat loss = 0 --> where ice melting (Q1) + icewater warming (Q2) (being the heat gained) and the heat loss being water cooling (Q3)
So I made these equations:
Q1: Heat of fusion water * mass of ice = 334400 * x = 334400x
Q2: Specific heat capacity for ice* m * ∆T = 2097 * 67 * x = 140499x
Q3: 20 * 4180 * -43 = -3594800J
Q1+Q2+Q3 = 0
334400x + 140499x - 3594800 = 0

474899x = 3594800
/474899 = /474899
x (mass of ice) = 7,57 kg
So I can't seem to get the correct answer.. is my ∆T for ice wrong?
Appreciate any help! :)

Think of all the steps that happen: The ice is warmed, then it melts, then the resulting water is warmed.
The heat to do these 3 things comes from the surrounding water, which cools off in the process.
Analyze the problem before you calculate.
See also the other answers you got!

 

[Polarbear10]

I took $Q_w$ to be the heat lost by the water:$$Q_w = M\Delta T_3 c_w$$where $M = 20 \ kg$, $\Delta T_3 = 43 \ K$ and $c_w$ is the specific heat of water.

Then, I took the mass of ice to be $m$ and calculated the three stages of heating:
$$Q = Q_1 + Q_2 + Q_3 = mq_1 + mq_2 + mq_3$$Then I equated the two and took out the common factor of $m$, so I ended up with:$$m = \frac{Q_w}{q_1 + q_2 + q_3}$$I find all those numbers get confusing. I prefer to let a computer do the arithmetic.

You need to find the correct expressions for the three stages of what started as ice being heated.

I just can't seem to get the correct result.. what numbers did you get? I bet it would be easier using a computer yes, I just really want to learn how to do it on paper before using a computer.

So ok, the ice goes from -12, freezing, and into melting at 0°C..
For Q1, I then used the Heat of fusion water * mass of ice = 334400 * x = 334400x

Q2: Then the icewater heats up from the surrounding water from 0°C and up to 55°C. Using Specific heat capacity for water* m * ∆T = 4180* 55* x = 229900x

Q3: The water already there goes from 98°C and cools down to 55°C, using Specific heat capacity for water* m * ∆T => 20 * 4180 * -43 = -3594800J
Dividing Q3 on Q1+Q2 gives me 6,37kg of ice..

So I must be doing something wrong, or if there is something I'm forgetting..

 

[Polarbear10]

Du brukte ikke feil nummer, du brukte feil tilnærming. Siden isvann og flytende vann ikke har samme varmekapasitet, må du dele prosessen i flere trinn.

Do I need to make a Q4, for the process when the ice goes from 0-12°C? Using Specific heat capacity for ice* m * ∆T=> 2097*12 = 25164J

Then in the end, add together Q1+Q2+Q4 / Q3(The heat lost)
Q1 334400 +Q2 229900 + Q4 25164 = 589464J
Q3 = 3594800J
3594800J/589464J = 6,01 kg

Am I getting this one right?

 

[DrClaude]

Do I need to make a Q4, for the process when the ice goes from 0-12°C? Using Specific heat capacity for ice* m * ∆T=> 2097*12 = 25164J

Then in the end, add together Q1+Q2+Q4 / Q3(The heat lost)
Q1 334400 +Q2 229900 + Q4 25164 = 589464J
Q3 = 3594800J
3594800J/589464J = 6,01 kg

Am I getting this one right?

I guess you made a typo there: it should be 6,1 kg (6,09 kg rounded up), because otherwise it is correct.

 

[Polarbear10]

I guess you made a typo there: it should be 6,1 kg (6,09 kg rounded up), because otherwise it is correct.

Yes, I got 6.098 and rounded it up to 6,1kg. But the process is correct?

What is the real difference between Q1 and Q4?
With Q1 being: Heat of fusion water * mass of ice = 334400 * x = 334400J
And Q4: Specific heat capacity for ice* m * ∆T=> 2097*x*12 = 25164J

Because isn't both of these the energy used to take the ice from -12°C --> 0°C?

 

[PeroK]

I just can't seem to get the correct result.. what numbers did you get? I bet it would be easier using a computer yes, I just really want to learn how to do it on paper before using a computer.

Okay, but your original post shows you drowning in a sea of numbers. You ought to treat algebra as a boat, to help you navigate the deep waters of university physics.

It's hard for me, as I have a particular inability to make sense of lines of digits. I really struggled to see what you were doing in your OP and spot your mistake.

Ultimately, however, your mistake seems to be to have forgotten that the ice needs to heat up from $-12$ degrees.

I can't see any real point in doing multiple long multiplications by hand in this day and age.

 

[Polarbear10]

Okay, but your original post shows you drowning in a sea of numbers. You ought to treat algebra as a boat, to help you navigate the deep waters of university physics.

It's hard for me, as I have a particular inability to make sense of lines of digits. I really struggled to see what you were doing in your OP and spot your mistake.

Ultimately, however, your mistake seems to be to have forgotten that the ice needs to heat up from $-12$ degrees.

I can't see any real point in doing multiple long multiplications by hand in this day and age.

You're right, however on tests and on exams we are only allowed to use pen and papir and calculator. No computers.

 

[jbriggs444]

You're right, however on tests and on exams we are only allowed to use pen and papir and calculator. No computers.

Then the smart thing to do is to minimize calculation. Use algebra to get a formula. Then use the calculator to evaluate the formula.

 

[PeroK]

You're right, however on tests and on exams we are only allowed to use pen and papir and calculator. No computers.

I don't know what marking schemes are employed, but I would be worried about submitting a page of numbers with the wrong number at the end and leaving it to the examiner to sift through the wreckage.

Instead, a well-thought out solution with a mistake in the final calculations ought to count for more.

I'd speak to someone in your physics department about whether you should be moving away from the "plug and chug" to a more mature, algebraic approach to physics problems. See what they say.

 

[kuruman]

I'd speak to someone in your physics department about whether you should be moving away from the "plug and chug" to a more mature, algebraic approach to physics problems. See what they say.

Amen to that. Students should spend their allotted test time thinking and instructors should evaluate understanding, not calculator prowess.