Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
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Chapter 13, Problem 56AE
Interpretation Introduction

Interpretation:

Energy needed for conversion of 100 g of ice to water at 20 °C has to be calculated.

Concept Introduction:

Heat required by 1 g of a substance to increase the temperature by 1 °C is termed as specific heat capacity of that substance. Energy absorbed or released by the substance is calculated as follows:

  Energy=(mass)(specific heat)(change in temperature)

Heat of fusion is amount of heat required to convert solid to liquid. Every substance has different heat of fusion. Heat of fusion for ice is 384 J/g. It is represented as ΔHfus and calculated as follows:

  q=(mass)(ΔHfus)

Expert Solution & Answer
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Answer to Problem 56AE

Energy needed for conversion of 100 g of ice to water at 20 °C is 48.78 kJ.

Explanation of Solution

The formula to calculate energy of ice is as follows:

  E=mc(T2T1)        (1)

Here,

E is energy of ice

m is mass of ice

c is specific heat of ice

T2T1 is change in temperature

Substitute 100 g for m, 2.01 J/g °C for c, 0 °C for T2 and 10.0 °C for T1 in equation (1).

  E=(100 g)(2.01 J/g °C)(0°C(10°C))=(100 g)(2.01 J1 g °C)(103 kJ1 J)(10 °C)=2.010 kJ

The formula to calculate energy needed for conversion of ice to water is as follows:

  q=mΔHfus        (2)

Here,

q is the energy needed for conversion of ice to water

m is the mass of ice

ΔHfus is the heat of fusion

Substitute 100 g for m and 384 J/g for ΔHfus in equation (2).

  q=(100 g)(384 J1 g)(103 kJ1 J)=38.400 kJ

The formula to calculate energy of water is as follows:

  E'=mc(T2T1)        (1)

Here,

E' is energy of water

m is mass of water

c is specific heat of water

T2T1 is change in temperature

Substitute 100 g for m, 4.184 J/g °C for c, 20 °C for T2 and 0 °C for T1 in equation (1).

  E'=(100 g)(4.184 J/g °C)(20°C(0°C))=(100 g)(4.184 J1 g °C)(103 kJ1 J)(20 °C)=8.368 kJ

Therefore, total energy needed for conversion of ice to water is as follows:

  Total energy of ice=E+q+E'        (3)

Substitute 2.010 kJ for E, 38.400 kJ for q and 8.368 kJ for E' in equation (3).

  Total energy of ice=2.010 kJ+38.400 kJ+8.368 kJ=48.78 kJ

Hence, energy needed for conversion of 100 g of ice to water at 20 °C is 48.78 kJ.

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Chapter 13 Solutions

Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card

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