   Chapter 10, Problem 101E

Chapter
Section
Textbook Problem

What quantity of energy does it take to convert 0.500 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g • °C; liquid, 4.2 J/g . °C; steam, 2.0 J/g • °C; ∆Hvap = 40.7 kJ/mol; ∆Hfus = 6.02 kJ/mol.

Interpretation Introduction

Interpretation: The required amount of heat for water at -20°C to 250°C should be calculated.

Concept Introduction:

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

C = Heat absorbedTemperature change

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

Specific heat capacity =  Absorbed heat (J)/ Temperature change(c)×mass ofsubstance (g)......(1)

For the above equation heat is:

q = S×M×ΔT......(2)

q is heat J

M is mass of sample (g)

S is specific heat capacity (J/°C·g)

ΔT  is temperature change (°C)

Explanation

Explanation

Record the given data:

Specific heat capacity of ice =   2.03 J/C.g

Specific heat capacity of steam = 2.0 J/C.g

Specific heat capacity of water 4.184 J/C.g

The  ΔΗvap =40.7 kJ/mol

The ΔΗfus = 6.02 kJ/mol

• The given specific heat capacities, ΔΗvap and ΔΗfus are recorded as shown above.

To calculate the required heat of ice -20°C to 0°C .

The reaction is,

H2O(s-20°C)H2O(s0°C)

q1=2.30Jg.°C×5.00×102×20°C=2.0×104J=20kJ

• The specific heat capacity of ice, temperature change and mass of ice are plugging in to equation 2 to give heat of ice -20°C to 0°C .
• The heat of ice -20°C to 0°C is 20kJ .

To calculate the required heat of ice 0°C to water at 0°C .

The reaction is,

H2O(sol0°C)H2O(lq0°C)

Molecular weight of water is 18.02g

q2=×5.00×102×1mol18.02g×6.02kJmol=167kJ

• The enthalpy of fusion and molecular weight of water are plugging in to equation  to give heat of ice 0°C to water at 0°C .
• The heat ice 0°C to water at 0°C is 167kJ .

To calculate the required heat of water at 0°C in to 100°C .

The reaction is,

H2O(lq0°C)H2O(1q100°C)

q3=4.2Jg.°C×5.00×102×100°C=2

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