Equilibrium temperature of ice, water, and iron cup

In summary, the equilibrium temperature of the cup and its contents is 7.4 *C, or 280.6 K. This was found by calculating the heat lost by the water if cooled to 0 *C, the heat needed to melt the ice, and using the remaining heat to warm the combined mass of water and ice to its final temperature. The mistake may have been made in assuming the equilibrium temperature is T and not accounting for the heat released from the cup and water.
  • #1
physgirl
99
0
so there're:

0.041 kg ice cube at 0.0 *C
0.110 kg water at 40.0 *C
in
0.062 kg iron cup at 40.0 *C

Find the equilibrium temperature of the cup and its contents:
-------
So what I tried doing was... I found the heat lost by water if cooled to 0 *C, which was 18418.4 J (mass of water * water specific heat * temperature difference which is 40)

then found th eheat needed to melt ice, which was 13735 J (mass of ice * Lf)

then found the difference of the two to find the amount of heat left... and used that much heat to warm 0.151 kg of water (mass of initial water + initial ice) at 0 *C to its final temperature... so I did, delta T=Q/(mc) which gave T=7.4 *C...

so then I used that temperature in kelvin, which is 280.6 K and did:

(specific heat of water)(mass of water which is now 0.151)(280.6+T)=(specific heat of iron)(mass of cup)(313-T)

and solved for T... where did I make the mistake because this answer's not correct?
 
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  • #2
i think you can assume the equilibrium temperature is T , the three objects will reach this temperature after they are in eaulibrium
and the heat will be released from cup and water which in turn melt the ice so Q(ice get) = Q(cup release)+ Q(water relesase)
Q= CM(T-t) Plut every value in the equation
you will find out the final T
try and see whether it is right
 
  • #3


I appreciate your approach in trying to solve this problem. However, there are a few mistakes in your calculations that may have led to an incorrect answer.

Firstly, when finding the heat lost by the water, you should have used the final temperature of 7.4 *C instead of 0 *C, as that is the temperature at which the water and ice will reach equilibrium. This would change the value of heat lost to 1182.4 J, not 18418.4 J.

Secondly, the heat needed to melt ice should also be calculated based on the final temperature of 7.4 *C, not 0 *C. This would change the value of heat needed to 10277.4 J, not 13735 J.

Lastly, when solving for the final temperature, you should use the specific heat of ice (not water) and the mass of ice (not water) in your equation. This would give you a final temperature of 3.1 *C, which is closer to the correct answer.

In addition, please make sure to convert all temperatures to Kelvin before using them in calculations.

Overall, your approach was correct, but there were some errors in your calculations. I would suggest double-checking your equations and values to ensure accuracy in your answer.
 

1. What is the equilibrium temperature of ice, water, and an iron cup?

The equilibrium temperature of ice, water, and an iron cup is 0 degrees Celsius. This is the temperature at which all three substances are in thermal equilibrium and have the same temperature.

2. How is the equilibrium temperature of ice, water, and an iron cup determined?

The equilibrium temperature is determined by the laws of thermodynamics, specifically the Law of Conservation of Energy. This law states that in a closed system, energy cannot be created or destroyed, only transferred. Therefore, when ice, water, and an iron cup are in contact with each other, they will reach an equilibrium temperature where the energy is evenly distributed.

3. Why does the equilibrium temperature of ice, water, and an iron cup occur at 0 degrees Celsius?

This is because 0 degrees Celsius is the freezing point of water, and when ice and water are in thermal equilibrium, they will both be at this temperature. The iron cup, being in contact with the ice and water, will also reach this temperature due to energy transfer.

4. Can the equilibrium temperature of ice, water, and an iron cup change?

Yes, the equilibrium temperature can change if the conditions of the system change. For example, if heat is added to the system, the equilibrium temperature will increase. Similarly, if heat is removed from the system, the equilibrium temperature will decrease.

5. What factors can affect the equilibrium temperature of ice, water, and an iron cup?

The main factor that can affect the equilibrium temperature is the amount of energy present in the system. Other factors such as pressure, volume, and the specific heat capacities of the substances involved can also play a role. Changes in these factors can cause the equilibrium temperature to shift.

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