   Chapter 11, Problem 22P

Chapter
Section
Textbook Problem

A 1.50-kg iron horseshoe initially at 600°C is dropped into a bucket containing 20.0 kg of water at 25.0°C. What is the final temperature of the water-horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

To determine
The final temperature of water-horseshoe system.

Explanation

Given Info: Initial temperature of water is 25.0°C , mass of water is 20.0 kg, initial temperature of the horseshoe is 600°C , and mass of horseshoe 1.50 kg.

Heat gained by the water is heat lost by the horseshoe.

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,

Horseshoe is made up of iron.

Formula to calculate Heat lost by horseshoe is,

QFe=mFecFe(TfTiFe)

• QFe is the heat lost by horseshoe,
• mFe is the mass of horseshoe,
• cFe is the specific heat of iron,
• TiFe is the initial temperature of horseshoe,

Heat gained by the water is heat lost by the aluminum QW=QFe .

Equate both heat energy equations and rewrite it in terms of m.

mWcW(TfTiW)=mFecFe(TfTiFe)mWcWTfmWcWTiW=mFecFeTf+mFecFeTiFe(mWcW+mFecFe)Tf=mWcWTiW+<

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