   # The complete combustion of acetylene, C 2 H 2 ( g ), produces 1300. kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal water by 10.0°c if the process is 80.0% efficient? Assume the density of water is 1.00 g/cm 3 • ### 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 7, Problem 90E
Textbook Problem
995 views

## The complete combustion of acetylene, C2H2(g), produces 1300. kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal water by 10.0°c if the process is 80.0% efficient? Assume the density of water is 1.00 g/cm3•

Interpretation Introduction

Interpretation: The amount of acetylene burned for the 80.0% efficient process should be calculated when the temperature raise 1.00 gal water by 10.0°C .

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 absorbed/ temperature change

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

Specific heat capacity =  Absorbed heat (J)/ Temperature change(c)×mass ofsubstance (g)...(1)

For the above equation heat is:

q = S×M×T....(1)

q is heat (J)

M is mass of sample (g)

S is specific heat capacity (J/°C·g.)

T is temperature change (C)

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

(absorbed)S×M×ΔT =-S×M×ΔT (released).....(3)

### Explanation of Solution

Explanation

Given data:

Process efficiency is 80%.

Molecular weight of acetylene 26.04g

Per mole of acetylene produced energy is 1300. kJ

Temperature change is 10.0°C .

Produced water is 1.00 gal.

To calculate: mass of water.

=1.00gal×3.785LL×1000mLL×1.00gmL=3790g

• The 1.00 gallons is 3.785 L so the given volume of water is calculate as gram by above method to get mass of water is 3790g .

To calculate: The energy (heat) required for 80.0% efficient process.

For the equation (3) the released heat is equal to absorbs heat of water

=4.18J/°C.g×3790g×10.0°C=1.58×105Jforthe80%precess=1

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