Chapter 10, Problem 101E

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404

Chapter
Section

### Chemistry

10th Edition
Steven S. Zumdahl + 2 others
ISBN: 9781305957404
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:

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

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Specific heat capacity of water 4.184Â J/C.g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  TheÂ  Î”Î—vapÂ =40.7Â kJ/mol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â The Î”Î—fusÂ =Â 6.02Â kJ/mol

• The given specific heat capacities, Î”Î—vap and Î”Î—fus are recorded as shown above.

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

The reaction is,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  H2O(s-20Â°C)â†’H2O(sâ€‰0Â°C)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  q1=â€‰2.30â€‰Jg.Â°CÃ—5.00Ã—102Ã—20Â°Câ€‰â€‰â€‰â€‰â€‰â€‰=â€‰2.0Ã—104Jâ€‰â€‰â€‰â€‰â€‰â€‰=â€‰20â€‰kJ

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

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

The reaction is,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  H2O(solâ€‰0Â°C)â†’H2O(lqâ€‰0Â°C)

Molecular weight of water is 18.02â€‰g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  q2=Ã—5.00Ã—102Ã—1â€‰mol18.02â€‰gÃ—6.02â€‰kJmolâ€‰â€‰â€‰â€‰â€‰â€‰=167â€‰kJ

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

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

The reaction is,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  H2O(lqâ€‰0Â°C)â†’H2O(1qâ€‰â€‰100Â°C)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  q3=â€‰4.2â€‰Jg.Â°CÃ—5.00Ã—102Ã—100Â°Câ€‰â€‰â€‰â€‰â€‰â€‰=â€‰2

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