   Chapter 10, Problem 75AP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
3 views

# A 25.0-g sample of pure iron at 85 °C is dropped into 75 g of water at 20. °C. What is the final temperature of the water−iron mixture?

Interpretation Introduction

Interpretation:

The final temperature of the water-iron mixture should be calculated.

Concept Introduction:

The specific heat is defined as the amount of heat per unit mass needed to raise the temperature by one degree Celsius.

It is mathematically represented as follows:

Q=msΔT

Here, Q is heat, m is mass, s is specific heat capacity and ΔT is change in temperature.

Explanation

From table 10.1, the specific heat capacity of water (liquid) is 4.184 J/g C

Also, the specific heat capacity of iron is 0.45 J/g C

The heat required can be calculated using the following equation:

Q=msΔT

Or,

Q=ms(T2T1)

Putting the values for iron:

Q1=(25.0 g)(0.45 J/g C)(T285 C)

Similarly, for water:

Q2=(75 g)(4.184 J/g C)(T220 C)

On adding both, the temperature of iron decreases and that of water increases, thus, the change is same but in opposite direction and, Q1=Q2

Or,

Q1+Q2=0

Putting the values,

(25.0 g)(0.45 J/g C)(T285 C)+(75 g)(4

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