Solving Heat Transfer Problem with Melting Ice Cube

In summary, the conversation is about a heat transfer problem involving a metal container, water, and an ice cube. The initial temperatures of the water and container are 22 degrees Celsius, while the ice cube is at 0 degrees Celsius. When thermal equilibrium is reached, the temperature of the system is 15 degrees Celsius. The equation for heat transfer in an isolated system is Q(water) + Q(ice) + Q(container) = 0. The confusion lies in whether or not to include the heat of transformation (Q=ML) in the equation for Q(ice). The experts advise focusing on the internal energy (U) and its change in the initial and final states of the system. They also discuss the heat
  • #1
david98999
17
1
Moved from a technical forum, so homework template missing
Good day members of physics forums .

I am a university student currently studying thermodynamics and I just am a bit confused about a heat transfer problem.

If you have the following items: 1 metal container which is well insulated from the outside ;
; 1 quantity of water and an icecube .

The water and the metal container are at 22 degrees celcius .
The ice cube is at the melting point 0 degrees celcius

The ice is added to the metal container and when thermal equilibrium is reached the temperature of the system is 15 degrees.

the heat transfer equation is Q(water)+Q(ICE) +Q(container)=0 in an isolated system where the heat of the container/water increases the temperature of the ice as the container/water decrease in temperature.

now with regards to the heat tranfer of the ice Q(ICE) I am a bit confused

I believe that the equation should be Q(ICE)=change in temperature =mc(Delta T)

but I am unsure if I should add the heat of transformation equation Q=ML .I think we don't need to because the ice cube is already at room temperature but i would kindly like to verify this because I am unsure.

Thank you .
 
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  • #2
Hi david:

I confess I did not follow all of you reasoning. However, I think you omitted the energy needed to melt the ice.

Hope this helps.

Regards,
Buzz
 
  • #3
What the heck is Q supposed to be in your equations? I though you said you are currently studying thermodynamics. If so, then you know you should be focusing on the internal energy U of the system in its initial and final states. Do you know what internal energy is, and how to calculate the change in internal energy?
 
  • #4
Chestermiller said:
What the heck is Q supposed to be in your equations? I though you said you are currently studying thermodynamics. If so, then you know you should be focusing on the internal energy U of the system in its initial and final states. Do you know what internal energy is, and how to calculate the change in internal energy?
-----------

My apologies . Q refers to the heat of transformation . I said that the net heat in this internal ice/container/water system is equal to zero .My textbook is Physics for scientists and engineers . Randall Knight 3rd edition
 
  • #5
OK. Let ##m_i##, ##m_w##, and ##m_c## be the initial masses of ice, water, and the container. Let the arbitrary reference state of zero internal energy (per unit mass) for ice and water be liquid water at 0 C, and the arbitrary reference state of zero internal energy (per unit mass) for the container be 0 C. So the initial internal energies per unit mass of the ice, water, and container are

$$u^0_i=-L$$
$$u^0_w=C_w(22-0)$$
$$u^0_c=C_c(22-0)$$
where the C's are the heat capacities of water and container, and L is the latent heat of melting ice.

Based on these relationships, what is the total internal energy ##U^0## of the system in its initial state?

In the final state of the system, all the ice is melted, and the total mass of liquid water is now ##(m_i+m_w)##, while the final temperature of the system is 15 C. Based on this, what is the final internal energy per unit mass of the water and of the container, ##u_w## and ##u_c##? What is the total internal energy of the system in its final state U?

Based on the first law of thermodynamics, how are U and ##U^0## related?

Chet
 

1. How does the melting of an ice cube affect heat transfer?

When an ice cube melts, it absorbs heat from its surroundings. This heat is used to break the bonds between the water molecules in the ice, causing the ice to melt and turn into liquid water. This process of absorbing heat is known as endothermic heat transfer.

2. What factors influence the rate at which an ice cube melts?

The rate at which an ice cube melts depends on several factors, including the temperature of the surrounding environment, the size and shape of the ice cube, and the type of material the ice cube is in contact with. For example, an ice cube will melt faster in a warmer room compared to a colder one, and it will melt faster on a metal surface compared to a wooden surface due to differences in conductivity.

3. How can the melting of an ice cube be used to solve heat transfer problems?

The melting of an ice cube can be used to calculate the amount of heat being transferred in a system. By measuring the change in temperature of the surrounding environment and the amount of time it takes for the ice cube to melt, the rate of heat transfer can be calculated using equations such as Q = mCΔT. This information can then be used to solve various heat transfer problems.

4. Can the melting of an ice cube be used to cool a room?

Yes, the melting of an ice cube can be used to cool a room through the process of evaporative cooling. As the ice cube absorbs heat from the surrounding environment and melts, the resulting water will evaporate into the air, taking some of the heat with it. This can create a cooling effect in the room, but it is not as efficient as using an air conditioning unit.

5. How does the melting of an ice cube compare to other methods of heat transfer?

The melting of an ice cube is just one of many methods of heat transfer. It is unique in that it involves a phase change from solid to liquid, but it follows the same principles as other forms of heat transfer, such as conduction, convection, and radiation. It is often used in heat transfer experiments and calculations because it is a simple and easily observable process.

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