Thermodynamics question regarding latent heat of fusion of ice.

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving the latent heat of fusion of ice. The scenario includes a copper container with water and ice, where the system reaches thermal equilibrium after the ice is added. Participants are exploring how to calculate the heat released by the container and water as they cool down, and how to determine the latent heat of fusion from this information.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of heat lost by the copper container and water, questioning how to incorporate these values into the overall energy balance. There are attempts to relate the heat lost to the mass and temperature change of the materials involved.

Discussion Status

Some participants have provided guidance on the equations to use for calculating heat loss and the relationship between the heat lost by the copper and water. However, there is no explicit consensus on the correct approach to find the latent heat of fusion, as different interpretations and calculations are being explored.

Contextual Notes

The original poster expresses difficulty in obtaining the correct value for the latent heat of fusion, indicating potential confusion regarding the application of the relevant equations and the energy balance in the system.

Bigdutchman
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Homework Statement


(a) An insulated copper container of mass 0.250 kg contains 0.350 kg water. Both the container and the water are initially at 25.0 0C. Then 0.012 kg of ice at 0.0 0C is added to the container. Eventually the container and contents reach thermal equilibrium at 21.7 0C.

(i) What is the total heat released (in J) by the copper container and the 0.350 kg water as they cool down from 25.0 0C to 21.7 0C?

I got 5154.104J (Correct answer)


(ii) Determine the latent heat of fusion of ice. <---- Can't do this mofo. Keep getting different answer. They have 3.39x10^5 J/kg. Can someone pls show me how they got that? THANKS!



Homework Equations


Q=mc delta T
Q=mL


The Attempt at a Solution


One of my failed attempts:
5154/0.012 = 4.29x10^5 J/kg
 
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You are not taking into account the heat lost by the copper and the water.

AM
 
how do i do that? lol
 
Bigdutchman said:
how do i do that? lol
If the copper changes temperature by \Delta T how much heat does it lose? Do the same for water.

AM
 
Q = mL for state changes. (mass, L being the heat needed, which is what you're trying to find)

and for heating of liquids, Q = mC * change in temp

heat lost by copper = -heat lost by water (i.e. heat is gained by water)
mC * change in temp for copper = - (mL + mC * change in temp for water)

rearrange to find L
 

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