   Chapter 11, Problem 26P

Chapter
Section
Textbook Problem

An unknown substance has a mass of 0.125 kg and an initial temperature of 95.0°C. The substance is then dropped into a calorimeter made of aluminum containing 0.285 kg of water initially at 25.0°C. The mass of the aluminum container is 0.150 kg, and the temperature of the calorimeter increases to a final equilibrium temperature of 32.0°C. Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.

To determine
The specific heat of unknown sample.

Explanation

Given Info: Initial temperature of water and aluminum calorimeter is 25.0°C , mass of aluminum calorimeter is 0.150 kg, mass of water is 0.285 kg, mass of unknown substance is 0.125 kg, initial temperature of the unknown substance is 95°C and final temperature is 32.0°C .

The heat lost by the unknown substance 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 unknown substance is,

QX=mXcX(TfTiX)

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

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

QW+QAl=QX

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

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