# Specific & Latent Heat - Heat required to turn water in Aluminum Tray -> Ice

• format1998
In summary: I'm not sure what is going wrong.I changed the sign of the heat removed to change phase (liquid to solid)to negative, and my answer came out to be exactly 35.1324 kJ. I even double checked it on a calculator.In summary, the amount of heat required to turn 200 g of water in an aluminum ice tray with a mass of 340 g from 18°C to -15°C is 35.1324 kJ.

#### format1998

Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

## Homework Statement

200 g Water (l) is contained in an Aluminum ice tray that has a mass of 340 g. Both is at 18°C. How much heat (Q) must be removed to turn the water into ice at -15°C?

Aluminum mAl= 340g
Water mW= 200g
Ti= 18°C

Tf= -15°C

## Homework Equations

Specific Heat Q = mcΔT
Latent Heat Q = mLf

## The Attempt at a Solution

QNET= heat removed to bring Aluminum from 18°C to -15°C
+ heat removed to bring Waterl from 18°C to 0°C
+ heat removed from water to change phase (liquid to solid)
+ heat removed to bring ice from 0°C to -15°C

QNET = (mcΔT)Al + (mcΔT)W(l) + (mLf)w + (mcΔT)W(s)

QNET = [.34kg (900 $\frac{J}{kg*°C}$)(-15°C-18°C)] + [.2kg (4186 $\frac{J}{kg*°C}$)(0°C - 18°C)] +[.2kg (333*103 $\frac{J}{kg}$)] + [.2kg (2100 $\frac{J}{kg*°C}$) (-15°C-0°C)
QNET = 35132.4 J = 35.1324 kJ

Thank you in advance. Any and all help is much appreciated!

when I did it I got 97.9kJ. I can only imagine that something went wrong with all the - signs.
For the Al I used temp change = 33 (18 to -15), for the water 18 and for the ice 15.

I think you might omit 103 from the latent heat.

ehild

ehild said:
I think you might omit 103 from the latent heat.

Latent Heat as given on the table was 333 kJ/kg. I needed it to be J/kg, which came out to be 333*103 J/kg.

There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.

technician said:
when I did it I got 97.9kJ. I can only imagine that something went wrong with all the - signs.
For the Al I used temp change = 33 (18 to -15), for the water 18 and for the ice 15.

"For the Al I used temp change = 33 (18 to -15)"

Q = mcΔT
33 grams for the mass of the Aluminum tray?
QAl= (.34 kg) (900 $\frac{J}{kg*C°}$)(-15°C - 18°C)
did I lay that out wrong or did I use the wrong numbers?

"for the water 18"
Water in liquid form is from 18°C to 0°C

"ice 15"
Ice from 0°C to -15°C

is that what you meant?

technician said:
There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.

Q(aluminum) = -10098 J
Q (water from 18°C to 0°C) = - 150696 J
Q (water to ice phase change) = 666000 J
Q (ice from 0°C to -15°C) = -6300 J

(-10098 J) + (-150696 J) + (666000 J) + (-6300 J) = 35,132.4 J = 35.13 kJ

What am I doing wrong? Are the signs wrong?

format1998 said:
QNET = [.34kg (900 $\frac{J}{kg*°C}$)(-15°C-18°C)] + [.2kg (4186 $\frac{J}{kg*°C}$)(0°C - 18°C)] +[.2kg (333*103 $\frac{J}{kg}$)] + [.2kg (2100 $\frac{J}{kg*°C}$) (-15°C-0°C)

The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

ehild

my values are
Q for aluminium =0.34 x 900 x 33 = 10098 J
Q water 18 to 0 = 0.2 x 4186 x 18 = 15070 J
Q water to ice = 0.2 x 333000 = 66600 J I think this is the difference!
Q ice to -15 = 0.2 x 2100 x 15 = 6300 J
These all add up to 10098 + 15070 +66600 +6300 = 98068 or 98kJ
We must be getting close ! I want to go to bed soon

Thank you both for your help! Much appreciated!

## 1. What is Specific Heat?

Specific heat is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius.

## 2. What is Latent Heat?

Latent heat is the amount of heat energy required to change the phase of a substance without changing its temperature.

## 3. How is Specific Heat different from Latent Heat?

Specific heat is the heat required to change the temperature of a substance, while latent heat is the heat required to change the phase of a substance.

## 4. Why is Specific Heat important in the context of turning water into ice?

Specific heat is important in this process because it determines the amount of heat energy required to lower the temperature of water to its freezing point and then to turn it into ice.

## 5. How does the material of the tray affect the heat required to turn water into ice?

The material of the tray, in this case aluminum, affects the heat required to turn water into ice because different materials have different specific heat capacities. Therefore, the amount of heat required to turn the water into ice will vary depending on the material of the tray.