# Solving Heat Transfer of Ice Homework: Initial Copper Temp

• Bassa
In summary: As such, your input may not be of much use to current users. In summary, the copper at 0C has a heat of fusion of 6390 J.
Bassa

## Homework Statement

A 6.00-kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.2 kg of ice and 0.80 kg of liquid water. What was the initial temperature od the piece of copper?[/B]

## Homework Equations

Q=mcΔT
Q=mL (L in this case is the heat of fusion)
ΣQ=0[/B]

## The Attempt at a Solution

Qcopper= 6(390)(Tf-T)
Qice=2(2100)(0-(-20))=8400J (heat required to raise the temperature of the ice to 0C)
Qwater=.8(334x10^3)=26400J (heat required to melt .8 kg of ice)

applying ΣQ=0:
6(390)(Tf-T) + 8400+26400=0

Tf= would have to be 0c because we still have some ice left.

solve for T:
T
=118.974C
what did I do wrong?[/B]

Last edited:

Chet

Chestermiller said:

Chet
Thanks! Is the method correct, though?

Bassa said:
Thanks! Is the method correct, though?
Yes

Bassa said:

## Homework Statement

A 6.00-kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.2 kg of ice and 0.80 kg of liquid water. What was the initial temperature od the piece of copper?[/B]

## Homework Equations

Q=mcΔT
Q=mL (L in this case is the heat of fusion)
ΣQ=0[/B]

## The Attempt at a Solution

Qcopper= 6(390)(Tf-T)
Qice=2(2100)(0-(-20))=8400J (heat required to raise the temperature of the ice to 0C)
Qwater=.8(334x10^3)=26400J (heat required to melt .8 kg of ice)

applying ΣQ=0:
6(390)(Tf-T) + 8400+26400=0

Tf= would have to be 0c because we still have some ice left.

solve for T:
T
=118.974C
what did I do wrong?[/B]

You should check your arithmetic here throughout. For example, when I calculate 2 * 2100 * 20, I don't get 8400.

Thank you very much for your help!

Your Latent heat of fusion of ice is incorrect. It should be 3.33 x 10^5 for it to have a result of 150 centigrade.

Fizixxs said:
Your Latent heat of fusion of ice is incorrect. It should be 3.33 x 10^5 for it to have a result of 150 centigrade.
Hello, @Fizixxs .

You are answering a thread which is more than 8 years old.

russ_watters

## 1. What is the purpose of solving heat transfer of ice homework?

The purpose of solving heat transfer of ice homework is to understand how heat is transferred between two objects, specifically in the context of ice and copper. This concept is important in many scientific fields, such as thermodynamics and materials science.

## 2. How do you calculate the initial copper temperature?

The initial copper temperature can be calculated using the formula: final temperature = (mass of ice x heat of fusion of ice) / (mass of copper x specific heat capacity of copper x change in temperature) This formula takes into account the heat absorbed by the ice as it melts and the heat lost by the copper as it cools.

## 3. What factors affect the heat transfer of ice to copper?

The heat transfer of ice to copper can be affected by several factors, including the initial temperature of the copper, the mass of the ice and copper, the specific heat capacities of the materials, and the shape and surface area of the objects. Environmental factors such as temperature and humidity can also impact heat transfer.

## 4. How does the rate of heat transfer change as the ice melts?

The rate of heat transfer from the ice to the copper will decrease as the ice melts. This is because the heat of fusion of ice (the energy required to melt ice into water) is much higher than the specific heat capacity of water. As the ice melts, it absorbs a significant amount of heat energy, slowing down the rate of heat transfer to the copper.

## 5. Can heat transfer of ice to copper be applied to real-life situations?

Yes, the concept of heat transfer between two objects is applicable to many real-life situations. For example, understanding heat transfer is important in designing and maintaining refrigeration systems, studying climate change, and developing new materials for various industries.

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