   # The freezing point of mercury is ―38.8 °C. What quantity of energy, in joules, is released to the surroundings if 1.00 mL of mercury is cooled from 23.0 °C to ―38.8 °C and then frozen to a solid? (The density of liquid mercury is 13.6 g/cm 3 . Its specific heat capacity is 0.140 J/g · K and its heat of fusion is 11.4 J/g.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5, Problem 21PS
Textbook Problem
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## The freezing point of mercury is ―38.8 °C. What quantity of energy, in joules, is released to the surroundings if 1.00 mL of mercury is cooled from 23.0 °C to ―38.8 °C and then frozen to a solid? (The density of liquid mercury is 13.6 g/cm3. Its specific heat capacity is 0.140 J/g · K and its heat of fusion is 11.4 J/g.)

Interpretation Introduction

Interpretation:

The energy released when mercury is cooled at different temperatures to its freezing point has to be calculated.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1k.Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where,

q= energy gained or lost for a given mass of substance (m),

C =specific heat capacity

ΔT= change in temperature

### Explanation of Solution

Given,

Freezing point of mercury=-38.80C

Volume =1mL

Density of mercury =13.6g/cm3

Specific heat capacity=0.140JK/g

Heat of fusion=11.4J/g

Mass =Density × volume

Substitute for the equation,

Mass=13.6g/cm3×1mL=13.6g

q1= Decrease the temperature from 23.00C to 00C

q=C×m×ΔT

Substitute for the equation as

q1=0

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