9th Edition
Steven S. Zumdahl
Publisher: Cengage Learning
ISBN: 9781133611097


9th Edition
Steven S. Zumdahl
Publisher: Cengage Learning
ISBN: 9781133611097


Chapter 6, Problem 58E
Textbook Problem

Hydrogen gives off 120. J/g of energy when burned in oxygen, and methane gives off 50. J/g under the same circumstances. If a mixture of 5.0 g hydrogen and 10. g methane is burned, and the heat released is transferred to 50.0 g water at 25.0ºC, what final temperature will be reached by the water?

Expert Solution
Interpretation Introduction

Interpretation: The final temperature water mixture to be determined.

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

  • Require heat of an 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 of substance (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)

                                                       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


given data:

                Energy given by hydrogen is 120. J/g

                 Energy given by methane is 120. J/g

                 Mass of water is 50.0 g

                 Initial temperature of water is 25.0 °C

                 Specific heat capacity of water is 4.18J /°C.g .

To determine: Final temperature of the water.

Energy given by hydrogen (qH) + Energy given by methane (qmethane) = Energy gained by water   (qw)...... (4)

      5.0 g ×120 J/g H+10 g×50 J/CH4 g = 1100 J                                             1100 J = 4.18J /C.g×50

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Chapter 6 Solutions

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