   Chapter 11, Problem 21P

Chapter
Section
Textbook Problem

An aluminum calorimeter with a mass of 0.100 kg contains 0.250 kg of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 50.0-g piece of copper at 80.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100 °C. The entire system stabilizes at a final temperature of 20.0°C. (a) Determine the specific heat of the unknown sample. (b) Using the data in Table 11.1, can you make a positive identification of the unknown material? Can you identify a possible material? (c) Explain your answers for part (b).

(a)

To determine
The specific heat of unknown sample.

Explanation

Given Info: Initial temperature of water and calorimeter is 10.0°C , mass of aluminum calorimeter is 0.100 kg, mass of water is 0.250 kg, mass of copper 50.0 g, mass of unknown block is 70.0 g, initial temperature of copper is 80.0°C , initial temperature of the unknown block is 100°C and final temperature is 20.0°C .

The heat lost by the copper and unknown block is equal to the heat gained by calorimeter and water.

Formula to calculate Heat gained by water is,

QW=mWcW(TfTiW)

• QW is the heat gained by water,
• mW is the mass of water,
• cW is the specific heat of water,
• TiW is the initial temperature of water,
• Tf is the final temperature of water,

Formula to calculate Heat gained by aluminum calorimeter is,

QAl=mAlcAl(TfTiAl)

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

Formula to calculate Heat lost by copper block is,

QCu=mCucCu(TfTiCu)

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

Formula to calculate Heat lost by unknown block is,

QX=mXcX(TfTiX)

• QX is the heat lost by unknown block,
• mX is the mass of unknown block,
• cX is the specific heat of the unknown block,
• TiX is the initial temperature of the unknown block,

The heat lost by the copper and unknown block is equal to the heat gained by calorimeter and water.

QW+QAl=(QCu+QX)

Use mAlcAl(TfTiAl) for QAl , mCucCu(TfTiCu) for QCu , mWcW(TfTiW) for QW , and mXcX(TfTiX) for QX to rewrite in terms of cX

(b)

To determine
The unknown material.

(c)

To determine
Explaining for part (b).

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