   Chapter 5, Problem 76GQ

Chapter
Section
Textbook Problem

Calculate the quantity of energy required to convert 60.1 g of H2O(s) at 0.0 °C to H2O(g) at 100.0 °C. The enthalpy of fusion of ice at 0 °C is 333 J/g; the enthalpy of vaporization of liquid water at 100 °C is 2256 J/g.

Interpretation Introduction

Interpretation:

The quantity of heat required to convert a definite mass of H2O(s) 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.

The heat of the substance can be calculated as:

q=heat of fusion or vaporization × mass

Explanation

To melt at 00C

The mass of the H2O is 60.1g

The heat of fusion of ice is 333J/g

q=heat of fusion × mass of H2O

q1=60.1g×333J/g=2.0×104J

To raise the temperature from 00C to 1000C

The mass of the water is 4.184J/gK

Mass of the water is 60.1g

The change in temperature is 100K

q2=4.184J/gK×60.1g×100K=2.5×104J

To evaporate at 1000C

The mass of the H2O is 60

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