   Chapter 11, Problem 20P

Chapter
Section
Textbook Problem

A large room in a house holds 975 kg of dry air at 30.0°C. A woman opens a window briefly and a cool breeze brings in an additional 50.0 kg of dry air at 18.0°C. At what temperature will the two air masses come into thermal equilibrium, assuming they form a closed system? (The specific heat of dry air is 1 006 J/kg · °C, although that value will cancel out of the calorimetry equation.)

To determine
At what temperature will the two air masses come into thermal equilibrium.

Explanation

Given Info: Mass of warm air is 975 kg, mass of cool air is 50.0 kg, initial temperature of the warm air is 30.0°C ,  initial temperature of the cool air is 18.0°C .

Formula to calculate Heat gained by cool air is,

QC=mCc(TfTiC)

• QC is the heat gained by cool air,
• mC is the mass of cool air,
• c is the specific heat of air,
• TiC is the initial temperature of cool air,
• Tf is the final temperature of cool air,

Formula to calculate Heat lost by warm is,

QW=mWc(TfTiW)

• QW is the heat lost by air,
• mW is the mass of warm air,
• TiW is the initial temperature of warm air,
• Tf is the final temperature of warm air,

Heat gained by the cold air is equal to heat lost by the warm air,

QC=QW

Use mCc(TfTiC) for QC   and mWc(TfTiW) for QW to rewrite in terms of Tf

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