   # A 110.-g sample of copper (specific heat capacity = 0.20 J/°C g) is heated to 82.4°C and then placed in a container of water at 22.3°C. The final temperature of the water and copper is 24.9°C. What is the mass of the water in the container, assuming that all the heat lost by the copper 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 58E
Textbook Problem
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## A 110.-g sample of copper (specific heat capacity = 0.20 J/°C g) is heated to 82.4°C and then placed in a container of water at 22.3°C. The final temperature of the water and copper is 24.9°C. What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?

Interpretation Introduction

Interpretation: The mass of water has to be calculated.

Concept Introduction: The heat capacity C is defined as the ratio of heat absorbed and temperature change. It can be given by,

C=heatabsorbed(q)temperaturechange(c)......(1)

• The required heat of one gram of substance raised to its temperature by one degree Celsius is called specific heat capacity.

Absorbed heat (q) = Specific heat capacity (S)×Temperature change(C)× mass ofsubstance (M)......(2)

From in above equation heat is:

S = q×M×ΔT.....(3)

Where,

S is specific heat capacity (J/°C·g)

q is heat (J)

M is mass of sample (g)

T is temperature change (ᵒC)

For any thermodynamic process, there is no loss of heat to the surrounding

Then the equation of heat can be given as,

(Absorbed) =-q (released)......(4)

### Explanation of Solution

Explanation

given data:

Mass of copper is 110g

Initial temperature of copper is 82.4 °C

Specific heat capacity of copper is 0.20 J/°C ·g

Final temperature of water mixture is 24.9°C

Initial temperature of water is 22.3°C

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

To determine: The mass of water.

Water gained heat is equals to heat lost by the copper so

(Absorbed)S×X×ΔT=S×X×ΔT (released)

4.18×X×(24.9-22.3) = 0

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