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Answer: A 5.00-g sample of aluminum pellets (specific heat

Chemical Principles | 8th Edition | ISBN: 9781305581982 | Authors: Steven S. Zumdahl ISBN: 9781305581982 176

Solution for problem 9.51 Chapter 9

Chemical Principles | 8th Edition

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Chemical Principles | 8th Edition | ISBN: 9781305581982 | Authors: Steven S. Zumdahl

Chemical Principles | 8th Edition

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Problem 9.51

A 5.00-g sample of aluminum pellets (specific heat capacity 5 0.89 J 8C21 g21) and a 10.00-g sample of iron pellets (specific heat capacity 5 0.45 J 8C21 g21) are heated to 100.08C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.08C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Step-by-Step Solution:

Problem 9.51

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J °C-1 g-1) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J °C-1 g-1) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Step by step solution

Step 1 of 2

When substance at lower temperature is mixed with a substance presenting at higher temperature then heat transfer from higher temperature substance to lower temperature takes place.

                   Heat lost by mixture of hot iron & aluminum = heat gained by water

                               .........(1)

                                                                  m =  mass

                                                                 C =  specific heat capacity

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Step 2 of 2

Chapter 9, Problem 9.51 is Solved
Textbook: Chemical Principles
Edition: 8
Author: Steven S. Zumdahl
ISBN: 9781305581982

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Answer: A 5.00-g sample of aluminum pellets (specific heat