Thermodynamics Problem(Water and Ice)

In summary, the conversation discusses a calorimeter experiment involving a 200g aluminum container and 500g of water at 20°C. A 100g piece of ice at -20°C is added to the system and the first question (a) asks for the final temperature of the system without any heat loss. The second question (b) involves adding another 200g piece of ice and determining the remaining amount of ice in the system at equilibrium. The specific heats of ice, water, and aluminum, as well as the equations Q=mcΔT and Q=mL, are provided as constants and equations to use in the calculations. The first question is answered to be 16.77°C and the second question is left
  • #1
timot0617
1
0
Two part question, I can answer the first one, but i don't know how

A 200g calorimeter made of aluminum contains 500g of water at 20°C. A 100g piece of ice is cooled to -20°C and is placed in the calorimeter.

(a)Find the final temperature of the system, assuming no heat loss.
(b)If we are to add a second 200g piece of ice at -20°C, how much ice remains in the system after it reaches equilibrium?

Some Constants and Equations
Specific Heat of Ice=2.0kJ/kg°C
Specific Heat of Water=4.186kJ/kg°C
Specific Heat of Aluminum=.91kJ/kg°C
Q=mcΔT and Q=mL

I answered the first one(a) and got 16.77°C.

I tried to answer the second one(b)

Q(1)=mcΔT=(.2kg)(2.0kJ/kg°C)(0°C-(-20°C)) [From -20°C to 0°C]
Q(2)=mL=(.2kg)(334kJ/kg)=66.8kJ [Ice-To-Water]

I'm stuck here... Also, am I doing it right?

Thanks for any help.
 
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  • #2
timot0617 said:
I answered the first one(a) and got 16.77°C.
Show how you got this answer.
 

FAQ: Thermodynamics Problem(Water and Ice)

How does thermodynamics explain the melting of ice into water?

Thermodynamics explains the melting of ice into water through the process of heat transfer. When heat is applied to ice, the molecules gain energy and start vibrating faster. This breaks the bonds holding the rigid ice structure together, causing it to melt into liquid water.

What is the relationship between temperature and the phase of water?

The phase of water (solid, liquid, or gas) is directly related to its temperature. At temperatures below 0°C, water exists as a solid (ice). Between 0°C and 100°C, it exists as a liquid. Above 100°C, water turns into a gas (steam).

How does the specific heat of water affect its thermodynamic properties?

The specific heat of water, which is the amount of heat needed to raise the temperature of 1 gram of water by 1°C, plays a crucial role in thermodynamics. Water has a high specific heat, meaning it can absorb a lot of heat energy without a significant change in temperature. This makes it a stable and important component in many thermodynamic processes.

Can ice and water exist in equilibrium at the same temperature?

Yes, ice and water can exist in equilibrium at the same temperature. This is known as the melting point (0°C) and occurs when the rate of melting equals the rate of freezing. At this point, the ice and water are in a constant state of flux, with molecules transitioning between the solid and liquid phases.

How do thermodynamic principles apply to the process of freezing water into ice?

The process of freezing water into ice is governed by the principles of thermodynamics. As heat is removed from liquid water, the molecules begin to slow down and form a rigid structure, forming ice. This release of heat energy is known as the latent heat of fusion, and it is the reason why ice feels cold to the touch.

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