   # Calculate the amount of energy necessary to raise the temperature of 1.00 L of ethanol ( d = 0.7849 g/cm 3 ) from 25.0 °C to its boiling point (78.3 °C) and then to vaporize the liquid. ( C ethanol = 244 J/g · K; heat of vaporization at 78.3 °C = 38.56 kJ/mol.) ### 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 5.3, Problem 1CYU
Textbook Problem
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## Calculate the amount of energy necessary to raise the temperature of 1.00 L of ethanol (d = 0.7849 g/cm3) from 25.0 °C to its boiling point (78.3 °C) and then to vaporize the liquid. (Cethanol = 244 J/g · K; heat of vaporization at 78.3 °C = 38.56 kJ/mol.)

Interpretation Introduction

Interpretation:

For a given volume of ethanol amount of heat energy required to raise the temperature to its boiling point and then to vaporize has to be determined.

Concept Introduction:

Heat capacity:

Heat energy required to raise the temperature of 1g of substance by 1k.

The amount of 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 data is:

Amount of ethanol is 1.0L

Specific heat capacity of ethanol is 2.44J/g.K

Heat of vaporization is 38.56kJ/mol

Density of ethanol is 0.7849g/cm3

Volume of ethanol is 1L

So, the mass of ethanol is 78.49kg

To raise the temperature of ethanol from 250C to 780C

q=C×m×ΔT

The relation between heat of vaporization and heat and mass can be represented as:

Hv

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