Specific Heat and Maybe Latent Heat also

[twilos]

Homework Statement

Several small piece of aluminum, having total (combined) mass of 2.5 kg are placed in liquid nitrogen and, when removed, are at a temperature of -135C. The aluminum is transferred to an insulating container with 0.50 kg of water. The water is at 25.0C [ Ignore heat lost from the system]

C Al = 910 J/kg K
C water = 4190 J/kg K ; Lf = 334000 J/kg ; Freezing pt = 0C
C ice = 2100 J/ kg K

Homework Equations

Q = mc(change)T
Lf = mLf
Conservation of Energy ( because system is isolated )
Qcold = -Qhot

The Attempt at a Solution

Before I attempted to do this problem, I wasnt sure if there would be a latent phase change for the water to ice because if there was then I would include that eq. Lf = mLf into my solution.

Gained Energy = Aluminum from water
Lose Energy = Water from Aluminum

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( T - 25C)

2275(T + 135C) + 2095 (T - 25C) = 0 ?

I am unsure if that's what your suppose to do and is it suppose to be +135C?

 

[rl.bhat]

You have to include Lf. So heat lost part include, heat lost by water from 25 degree to zero degree + heat lost by ice from zero degree to T degree.

 

[twilos]

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( 0 - 25C) +(x)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

Is this what your talking about? Because then there's 2 variables X because you don't know the amount of water that changes into ice...?

 

[rl.bhat]

To check whether the water completely convert in ice or not, calculate amount of heat absorbed by aluminum to reach 0 degree, and amount of heat lost by the water to convert water from 25 degree C to 0 degree C ice. If first one is greater than the second,water is completely converted into ice. Other wise a part of water is converted into ice.

 

[twilos]

that is necessary to find the temperature final? And if its partially converted doing what you said will give a ratio of how much ice is in the water?

 

[rl.bhat]

Yes.

 

[twilos]

okay so the following i did:

Qal = (2.5kg)(910 J/kg K) (135C) = 307125 J
Qw = (0.5kg)(4190 J/kg K) (25C) = 52375 J

Therefore the amount of heat for water is less than Aluminum so all of the water turns into ice?

then i can just do :

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( 0 - 25C) +(0.5 kg)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

and solve for T?

 

[rl.bhat]

Before that check the signs and parenthesis.

 

[twilos]

damn i can't get it to work out like i am really bad with the algebra lol anyone help me simplify this please?

 

[twilos]

(2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( T - 25C) +(0.5 kg)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

i believe all the numbers are correct however I am always confused because of the positives and negatives... and the left side is gaining heat and the right side is losing heat. So the left side has a Negative to begin with. Also the Aluminum is at -135 does that cancel out the - from Tf-Ti?

 

[rl.bhat]

To avoid confusion in signs, if heat is gained use Tf - Ti. If heat is lost, use Ti - Tf.
Then use heat gained = heat lost.