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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 200-g cube of copper that is initially at 85°C, and the other is a chunk of aluminum that is initially at 5.0°C. To the surprise of the student, the water reaches a final temperature of 25°C, precisely where it started. What is the mass of the aluminum chunk?

To determine
The mass of aluminum chunk.

Explanation

Given Info: mass of steel container is 120 g, mass of water is 150 g, mass of cube of copper is 200 g, Initial temperature of water is 25°C , Initial temperature of copper cube is 85°C , initial temperature of aluminum chunk is 5.0°C and final temperature is 25.0°C .

The temperature of water and steel container remains same and so their internal energy. Hence, the temperature lost by copper is gained by the aluminum and so the internal energy of copper decreases and the aluminum increases.

Formula to calculate Heat gained by aluminum is,

QAl=mAlcAl(TfAlTiAl)

  • QAl is the heat gained by aluminum chunk,
  • mAl is the mass of aluminum chunk,
  • cAl is the specific heat of aluminum,
  • TiAl is the initial temperature of aluminum chunk,
  • TfAl is the final temperature of aluminum chunk,

Formula to calculate Heat lost by copper is,

QCu=mCucCu(TfCuTiCu)

  • QCu is the heat lost by copper,
  • mCu is the mass of copper,
  • cCu is the specific heat of copper,
  • TiCu is the initial temperature of copper,
  • TfCu is the initial temperature of copper,

The heat lost by the copper and is equal to heat gained by Aluminum.

QAl=QCu

Use mAlcAl(TfAlTiAl) for QAl and mCucCu(TfCuTiCu) for QCu to rewrite in terms of mAl

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