Understanding Heat Transfer in a Ice-Water System

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SUMMARY

The discussion focuses on calculating the final temperature and the amount of ice melted in an insulated system containing an 18g ice cube at -4.0ºC and 75g of water at 10ºC. The relevant equations used include Q=MCΔT for heat transfer and Q=ML for phase changes, with specific heat capacities of ice (2.09 J/g°C) and water (4.186 J/g°C) provided. The calculations indicate that the final temperature of the system is approximately 9.517ºC, suggesting that not all ice melts due to insufficient heat energy from the water.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with specific heat capacity and latent heat concepts.
  • Proficiency in algebra for solving equations.
  • Knowledge of phase changes in substances, particularly ice and water.
NEXT STEPS
  • Study the concept of heat transfer in phase changes, focusing on latent heat calculations.
  • Learn about the first law of thermodynamics and its application in closed systems.
  • Explore more complex heat transfer scenarios involving multiple phases, such as ice to steam.
  • Practice solving problems involving calorimetry and energy conservation in thermal systems.
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone seeking to understand heat transfer in phase change scenarios.

KylieB
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Homework Statement



An 18g ice cube at -4.0ºC is placed into 75g of water at 10 ºC in an
insulated container.
a. What is the final temperature of the system?
b. How much of the ice melts?

I have a sheet full of these & ice to water to steam I really need someone to explain how to find the results so I can get my brain around these

Homework Equations


Q=MCΔT
C ice =2.09j/g°c
Cwater = 4.186j/g°c
M=Mass in g
T = temp in Degrees C

Q=ML
M=Mass in g
Lice =334j/g

The Attempt at a Solution


So here's what i think is right so far
Raise the temp of ice to 0
18 x 2.09 x 4 = 150 joules
Remove this energy from the water
150.48=75 x 4.186 x(10-Tf)is this part right?
150.48= 3139.5-313.95(Tf)
I am not sure how to get to the resultant temp of the water propery before I consider a phase change in the ice but I have a feeling there won't be enough heat energy in the water to melt all of the ice.
 
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So here's what i think is right so far
Raise the temp of ice to 0
18 x 2.09 x 4 = 150 joules
Remove this energy from the water
150.48=75 x 4.186 x(10-Tf)is this part right?
150.48= 3139.5-313.95(Tf)
I am not sure how to get to the resultant temp of the water propery before I consider a phase change in the ice but I have a feeling there won't be enough heat energy in the water to melt all of the ice
 
or to remove the energy is it simply 150.48 = 75 x 4.186 x Tf
150.48 = 315.95 (T?)
T? = 150.48/313.95
T? = 0.483degrees C
then remove this from the initial temp of water
10 - 0.483 = 9.517 Deg C is this right so far?
 

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