   # A 150.0-g sample of a metal at75.0ºC is added to 150.0 g H 2 O at 15.0ºC. The temperature of the water rises to 18.3ºC. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. ### 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 57E
Textbook Problem
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## A 150.0-g sample of a metal at75.0ºC is added to 150.0 g H2O at 15.0ºC. The temperature of the water rises to 18.3ºC. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

Interpretation Introduction

Interpretation: The specific heat capacity of the metal to 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

• Require heat of 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×mass ofsubstance (g)                          ......(1)

For in above equation heat is:

q = S×M×T......(2)q is heat (J)M is mass of sample (g)                                                     S is specific heat capacity (J/°C×g)                                                    T istemperature change (oC)

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

(absorbed) =-q (released)......(3)

### Explanation of Solution

Explanation

given data:

Mass of water = 150.0 gInitial temperature of water = 15.0 CFinal temperature of metal and water mixture = 18.3oCSpecific heat capacity of water is4.18J/oC.g.

To determine: Thetotal energy (heat) transfer in to water

heat given by metal (metal) = heat gained by water (qW)               (4)

4.18 × 150.0 × (18.3− 15.0)   =   2100 J

The given values are plugging in to equation (2) to get a total energy transfer in to water is 2100 J.

To determine: The specific heat capacity of the metal.

given data:

Mass of metal = 150

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