Thermodynamics - Hot copper block dropped in ice water

In summary, the experiment involves dropping a hot copper block into ice water, demonstrating the principles of thermodynamics, specifically heat transfer. The hot copper block loses thermal energy to the colder ice water, causing the temperature of the copper to decrease while increasing the temperature of the water and ice. This process illustrates key concepts such as conduction, specific heat capacity, and thermal equilibrium, where energy continues to flow until both substances reach a uniform temperature.
  • #1
nighfallraid
3
0
Homework Statement
A 0.20 kg block of copper is heated in a laboratory oven to a temperature of 180o
C. It is added to an insulated
container of ice and water which is at a temperature of 0o
C. The mass of ice is 0.10 kg and the mass of water
is 0.90 kg.
i) Calculate the heat required to melt all the ice in the container.
ii) Calculate the heat released if the 0.20 kg block of copper is cooled from 180o
C to 0o
C.
iii) Find the final temperature of the contents of the container.
Neglect the heat capacity of the container. Assume no heat is lost to the surroundings.
Relevant Equations
The specific heat capacity of copper is 381 J kg -1 K-1
The specific heat capacity of liquid water is 4.19 kJ kg -1 K-1
The latent heat of fusion of water is 333 kJ kg-1
The latent heat of vaporisation of water is 2260 kJ kg-1
1) Q=m*L
-.10*333
2)Q=mc deltat
-.10*381*180-0
im unsure about the rest
 
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  • #2
I think as a first step to part (iii) you want to figure out what temperature the copper would be if it melted all the ice.
 
  • #3
What are the units of the answers in parts 1 and 2
 
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Likes hutchphd and erobz
  • #4
Chestermiller said:
What are the units of the answers in parts 1 and 2
Chestermiller said:
What are the units of the answers in parts 1 and 2
kJ
 
  • #5
erobz said:
I think as a first step to part (iii) you want to figure out what temperature the copper would be if it melted all the ice.
can you show me how ?
 
  • #6
nighfallraid said:
2)Q=mc deltat
-.10*381*180-0
What is the mass of the copper?
Also it will release heat, so "heat released" should be positive.
nighfallraid said:
1) Q=m*L
-.10*333
Heat will need to be added to melt the ice, so "heat required" is positive.

For iii, you do not know at the start whether all the ice will melt. Just assume whatever seems the likelier of the two possibilities and calculate based on that.
If you get a silly answer (like, negative mass of ice left), switch to the other.
 
  • #7
nighfallraid said:
can you show me how ?
You know the heat required to be absorbed by the ice to melt it ##Q_{melt}##, you calculated it in (i). Equate that to the change in internal energy of the copper ball:

$$ \Delta U_c = - m_{ice}L $$
 
  • #8
nighfallraid said:
kJ
the units in (ii) are not ##\rm{kJ}##.
 
  • #9
What parts (i) and (ii) are begging at is for you to use them to determine if the final state has ice or not. In order for all the ice to be melted in the final state of equilibrium you have a certain amount of heat that needs to be absorbed by the ice before the supplier of said heat (the copper mass) reaches the temperature of the ice ( i.e. ##0~{}^{\circ}C## )., because heat will not be transferred without thermal gradient. Can the copper mass supply enough heat to do that?
 
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Likes Steve4Physics
  • #10
nighfallraid said:
kJ

The specific heat capacity of copper is 381 J kg -1 K-1
 

FAQ: Thermodynamics - Hot copper block dropped in ice water

What happens to the temperature of the copper block and the ice water when the hot copper block is dropped into the ice water?

When the hot copper block is dropped into the ice water, heat will transfer from the copper block to the ice water. As a result, the temperature of the copper block will decrease, while the temperature of the ice water will increase. If there is ice present, it may start to melt as it absorbs heat from the copper block.

Will the final temperature of the system be higher, lower, or the same as the initial temperature of the ice water?

The final temperature of the system will be higher than the initial temperature of the ice water. This is because the hot copper block transfers heat to the ice water, raising its temperature. However, the exact final temperature will depend on the masses and specific heat capacities of the copper and water, as well as the initial temperatures.

How do you calculate the final equilibrium temperature of the system?

To calculate the final equilibrium temperature, you can use the principle of conservation of energy. The heat lost by the copper block will be equal to the heat gained by the ice water. The equation is: \( m_{Cu} \cdot c_{Cu} \cdot (T_{initial, Cu} - T_{final}) = m_{H2O} \cdot c_{H2O} \cdot (T_{final} - T_{initial, H2O}) \), where \( m \) is mass, \( c \) is specific heat capacity, and \( T \) is temperature. Solve this equation for \( T_{final} \).

What role does the specific heat capacity play in this process?

The specific heat capacity determines how much heat is required to change the temperature of a substance by a certain amount. In this process, the specific heat capacity of copper and water will influence how much their temperatures change when they exchange heat. Water has a higher specific heat capacity than copper, meaning it will absorb more heat without a large change in temperature compared to copper.

What assumptions are made in the calculation of the final temperature?

Several assumptions are typically made in these calculations: (1) No heat is lost to the surroundings, meaning the system is perfectly insulated. (2) The copper block and water reach thermal equilibrium. (3) The specific heat capacities remain constant over the temperature range considered. (4) If ice is present, it melts completely before the water temperature rises above 0°C. These assumptions simplify the calculations but may not perfectly reflect real-world conditions.

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