Finding the Mass of an Aluminum Cup Based on Heat Transfer in a System

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Homework Help Overview

The problem involves calculating the mass of an aluminum cup based on heat transfer principles, specifically focusing on the thermal interactions between an ice cube and the cup as they reach thermal equilibrium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the heat transfer equations and the need to account for various phases of the ice as it melts, including the heat of fusion and temperature changes.

Discussion Status

Participants are exploring different aspects of the heat transfer calculations. Some have attempted to solve for the mass of the aluminum cup but have encountered incorrect results, prompting further examination of the necessary terms in the heat transfer equations.

Contextual Notes

There is a focus on ensuring that all relevant heat transfer components are included in the calculations, such as the heat required to change the temperature of the ice and the heat of fusion. Participants are working within the constraints of the problem's setup and the given temperatures.

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Homework Statement



A 105.05 g ice cube at -15°C is placed in an aluminum cup whose initial temperature is 73°C. The system comes to an equilibrium temperature of 24°C. What is the mass of the cup?

Homework Equations


Q=mc (delta T)


The Attempt at a Solution



the sum of total heat in the system is equal to zero.
Qice+Qaluminium cup=0
MiceCice(delta T)+MaluminiumCaluminium(delta T)=0
I solved for mass of aluminium cup, and got 0.1951 kg. But, it is wrong.
Please help.
 
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Since the ice melts, I think you need to take into account the heat of fusion, Q=mL, where m is the mass of the ice and L is the latent heat of fusion.

Try that.
 
I did, and got 0.8633 kg. But, that is wrong too.
 
You need to include terms for:
the heat required to bring the ice from -15 degrees to 0 degrees
the heat of fusion to melt the ice at 0 degrees to water at 0 degrees
the heat required to bring the water at 0 degrees to water at the final temperature of 24 degrees

This will be equal to the heat the aluminum cup loses.
 
Thank you very much.
 

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