College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 11, Problem 21P

(a)

To determine

The specific heat of unknown sample.

(a)

Expert Solution
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Answer to Problem 21P

The specific heat of unknown sample is 1822J/kg°C .

Explanation of Solution

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 .

mAlcAl(TfTiAl)+mWcW(TfTiW)=[mCucCu(TfTiCu)+mXcX(TfTiX)]mXcX(TfTiX)=mAlcAl(TfTiAl)+mWcW(TfTiW)+mCucCu(TfTiCu)

Substitute 0.250 kg for mW , 0.100 kg for mAl , 900J/kg°C for cAl , 20.0°C for Tf , 10.0°C for TiW , 10.0°C for TiAl , 4186J/kg°C for cW , and 10.0°C for TiW , 50.0 g for mCu , 387J/kg°C for cCu , 80.0°C for TiCu , 100.0°C for TiX , 70.0 g for mX to find cX .

[(70.0g)(1kg103g)cX(20.0°C100.0°C)]={(0.100kg)(900J/kg°C)(20.0°C10.0°C)+(0.250kg)(4186J/kg°C)(20.0°C10.0°C)+(50.0g)(1kg103g)(387J/kg°C)(20.0°C80.0°C)}(5.6kg°C)cX=10204JcX=1822J/kg°C

Conclusion:

Therefore, the specific heat of unknown sample is 1822J/kg°C .

(b)

To determine

The unknown material.

(b)

Expert Solution
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Explanation of Solution

The value of specific heat help to determine the substance.

Substance has their own value of specific heat. It is amount of heat required to raise the temperature of 1 kg of substance to 1°C . The calculate value of specific heat resembles the specific heat of Beryllium.

Conclusion: Therefore, the unknown substance May be Beryllium

(c)

To determine

Explaining for part (b).

(c)

Expert Solution
Check Mark

Explanation of Solution

If a person has more specific heat it requires more heat energy to raise its temperature. Specific heat helps in determining the substance.

Water has specific heat capacity of 4186J/kg°C , when common salt is added to the water, the specific heat of the Answer changes. So, the value for specific heat obtained in part (a) can be specific heat of a metal or alloy. Hence, the substance is not exactly identified with the specific heat.

Conclusion: The substance could be Beryllium, alloy or any other material.

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An aluminum calorimeter with a mass of 100 g contains 250 g 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 as shown, can you make a positive identification of the unknown material? Can you identify a possible material? (c) Explain your answers for part (b).
An aluminum calorimeter with a mass of 100 g contains 250 g 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. Determine the specific heat of the unknown sample and Using the data in Table of constants, identify of the unknown material?
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