   # In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at 1.00 × 10 4 Btu/h (1 Btu, or British thermal unit = amount of energy to raise the temperature of 1 lb water by 1°F). What quantity of water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 9, Problem 119AE
Textbook Problem
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## In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at 1.00 × 104 Btu/h (1 Btu, or British thermal unit = amount of energy to raise the temperature of 1 lb water by 1°F). What quantity of water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner?

Interpretation Introduction

Interpretation: The amount of evaporated water from electric air conditioner per hour should be calculated.

Concept Introduction:

The heat capacity C is defined as the ratio of heat absorbed to the temperature change. It can be given by,

C = Heat absorbedTemperature change......(1)

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

heat is:

q = S×M×ΔT......(2)

q is heat J

M is mass of sample (g)

S is specific heat capacity (J/°C·g)

ΔT  is temperature change (C)

For the process no heat loss to the surroundings means then the heat is

(absorbed) =-q (released).....(3)

### Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of water = 4.184 J/C.g

The electric air conditioner is rate = 1.00 X104 Btu/h .

• The given specific heat capacity and rate of electric air conditioner are recorded as shown above.

To convert the mass of water in gram and the amount of heat in J.

=1.00lb×454glb=454g

The gained heat is

=4.18Jg°C×45459°C=1.05×103J=1.05kJ

• The given rate of air conditioner and the specific heat capacity of water are plugging in the above equations to give the mass of water in gram and the amount of heat in J.
• of water in gram and the amount of heat in J.
• The mass of water is 454g and  heat gain is 1

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