Intro Thermodynamics equilibrium problem

AI Thread Summary
A 3.6-kg block of ice at 263 K is in thermal contact with a 13-kg block of silver at 1006 K, and the system is insulated. The equilibrium temperature will be determined by the heat exchange between the ice and the silver, considering the phase changes of the ice. The ice will first warm to 273.15 K, undergo a phase change to water, and then heat further to reach the final equilibrium temperature. Calculations indicate that the final temperature will be less than 100°C due to the heat capacity of the silver and the latent heat involved in the phase changes. The thread was closed as the original poster had not participated for over three years.
tri5
Messages
11
Reaction score
0

Homework Statement



A 3.6-kg block of ice originally at 263 K is placed in thermal contact with a 13-kg block of silver (cAg = 233 J/kg-K) which is initially at 1006 K. The H2O and silver system is insulated from any other heat transfer.

1)At what temperature will the system achieve equilibrium?
2)What will be the phase of the H2O at equilibrium?

CONSTANTS
g = 9.81 m/sec2
π = 4*arctan(1)
Atmospheric Pressure = 101,300 Pa
Density of Freshwater = 1000 kg/m3
Density of Saltwater = 1028 kg/m3
Specific Heat Capacity of Water = 4186 J/K*kg
Specific Heat Capacity of Ice = 2200 J/K*kg
Latent Heat of Fusion for Water = 333,400 J/kg
Latent Heat of Vaporization for Water = 2,260,000 J/kg
Gas Constant R = 8.134 J/mol-K
σb = 5.670 X 10-8W m-2K-4
k = 1.37 X 10-23 J/K
0°C = 273.15K

Homework Equations


Q=m*c*ΔT
Q=m*C

The Attempt at a Solution


In the attachment below.
As a note I used the specific heat of water(4186 J/K*kg) for C sub water,T sub i for water as 373.15 K and T sub i for silver as 1006 K.

I used the fact the heat from the block of silver is released and absorbed by the block of ice. The ice will have to raise its temperature of 263 K to 273.15 K, then it go through a phase change turning to water, then the waters temperature is raised to 373.15 K, then the water goes through another phase change to turn into steam, the final temperature of the steam is then the temperature of the system in thermal equilibrium. I made the previous assumptions due to the block of silver is at 1006 K which is 723 °C(way above the boiling point of water). The first time I worked out the problem I got a ridiculous answer(it came out to a figure that suggested the system increased in temperature). I then changed a few signs, which then gave me a more reasonable answer, that still gave me a wrong answer. I'm starting think that the ideal gas law may play a part in this, since the water has turned into steam.
 

Attachments

  • A68MS8hBHHELBdAB4EAR1T53.jpg
    A68MS8hBHHELBdAB4EAR1T53.jpg
    47.4 KB · Views: 1,220
Physics news on Phys.org
From your calculations, how much heat would you have to add to raise the temperature of the water from ice at -10 C to liquid water at 100 C?

How much heat would you have to remove from the silver to cool it from 1006 K to 373 K (100 C)?

What does this tell you about whether the final temperature is going to be >100C or <100C?

Chet
 
Thanks for the help! I figured it out. The ice cube goes from 263 K to 273.15 K, then a phase change from solid to liquid, then the liquid(water) increases in temperature to the final temperature,which is the temperature of equilibrium for the system.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top