Chemistry: The Science in Context (Fourth Edition)
Chemistry: The Science in Context (Fourth Edition)
4th Edition
ISBN: 9780393124187
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 5, Problem 5.102QP

(a)

Interpretation Introduction

Interpretation: The questions based on the fuel value and the heat energy is to be stated.

Concept introduction: A fuel is a substance which releases energy on combustion. The released energy may in the form of heat, light or may in other form. Fuel value is one of the basic characteristic of the fuel. It is defined as the amount of energy released by the fuel when 1g of the substance/fuel undergoes complete combustion. The main source of fuels is the hydrocarbon. The unit of fuel value is kJ/g . Heat capacity is the calculated amount of energy of the specific mass of the compound to raise the temperature in a fixed range.

To determine: The amount of fuel value for the compound C6H14 .

(a)

Expert Solution
Check Mark

Answer to Problem 5.102QP

Solution

The amount of fuel value for the compound C6H14 is 48.31kJ/g_ .

Explanation of Solution

Explanation

Given

The given amount of enthalpy of combustion ΔHcombustion° for the compound C6H14 is 4163kJ/mol .

The given compound C6H14 is a hydrocarbon. The major use of hydrocarbon in our daily life is as a fuel. A hydrocarbon generates energy on combustion with oxygen to generate electricity, to power vehicles, to cook our meals etc. The amount of enthalpy change during the combustion reaction is known as the ΔHcombustion° . As already defined above that the fuel value is the released amount of energy on complete combustion of per gram of the substance. For the above given hydrocarbons, fuel value is calculated by the formula,

Fuelvalue=ΔHcombustion°M (1)

Where,

  • ΔHcombustion° is the standard enthalpy of combustion.
  • M is the molar mass of the hydrocarbon.

The molar mass of the compound C6H14 is given as,

86.172g/mol

Substitute the values in the equation (1).

Fuelvalue=ΔHcombustion°MFuelvalue=4163kJ/mol86.172g/molFuelvalue=48.31kJ/g_

(b)

Interpretation Introduction

To determine: The amount of heat released during the combustion of 1.00kg of C6H14 .

(b)

Expert Solution
Check Mark

Answer to Problem 5.102QP

Solution

The amount of heat released during the combustion of 1.00kg of C6H14 is 48.31×103kJ_

Explanation of Solution

Explanation

As already explained above that the amount of heat released by the fuel when 1g of the substance/fuel undergoes complete combustion is known as fuel value. It means to calculate the amount of heat released during the combustion of 1.00kg of C6H14 the fuel value in terms of per kilogram is to be calculated. Therefore, the amount of heat released during the combustion of 1.00kg of C6H14 is the amount of fuel value for the combustion of 1.00kg compound and it is expressed as,

Fuelvalue=48.31kJ/gFor1.00gcompoundenergyreleased=48.31kJFor1.00kgcompoundenergyreleased=48.31×103kJ_

(c)

Interpretation Introduction

To determine: The amount of the compound C6H14 to heat 1.00kg of water from 25.0°C to 85.0°C .

(c)

Expert Solution
Check Mark

Answer to Problem 5.102QP

Solution

The amount of the compound C6H14 to heat 1.00kg of water from 25.0°C to 85.0°C is 5.189g_ .

Explanation of Solution

Explanation

The amount of heat (q) required to warm 1.00kg of water from 25.0°C to 85.0°C is calculated by the formula,

q=nCpΔT (1)

Where,

  • n is the number of moles of the substance.
  • Cp is the molar heat capacity of the substance.
  • ΔT is the change in temperature.

Molar mass of water is 18.02g/mol .

Molar heat capacity of water is 75.3J/mol°C .

For 1.00kg of water the number of moles (n) of water is calculated by the formula,

n=mM

Where,

  • m is the given mass of water.
  • M is the molar mass of water.

Substitute the values in the above equation.

n=mMn=1.00×103g18.02g/moln=55.49moles

Now, substitute the values in equation (1).

q=nCpΔTq=55.49moles×75.3J/mol°C(8525)°Cq=250703.82Jq=250.70kJ

This calculated value of heat capacity is the required amount of heat to warm 1.00kg of water from 25.0°C to 85.0°C .

As already calculated in part (a) that the amount of energy released by compound C6H14 on combustion of 1.00g is 48.31kJ . The negative sign shows the released amount of energy.

From these two calculations the amount of the compound C6H14 to heat 1.00kg of water from 25.0°C to 85.0°C is calculated as,

48.31kJheatobtained=1.00gC6H141kJheatobtained=1.00gC6H1448.31kJ250.70kJheatobtained=1.00gC6H1448.31kJ×250.70kJ250.70kJheatobtained=5.189g_

(d)

Interpretation Introduction

To determine: The amount of the white gas (25%C5+75%C6) to heat 1.00kg of water from 25.0°C to 85.0°C .

(d)

Expert Solution
Check Mark

Answer to Problem 5.102QP

Solution

The amount of the white gas (25%C5+75%C6) to heat 1.00kg of water from 25.0°C to 85.0°C is 5.17g_ .

Explanation of Solution

Explanation

According to the assumption the white gas has 25% C5 hydrocarbons and 75% C6 hydrocarbons. Therefore, the amount of hydrocarbons in calculated as,

For 25% C5 hydrocarbons,

1gWhitegascontains=25%hydrocarbon=25100ghydrocarbon=0.25ghydrocarbon

For 75% C6 hydrocarbons,

1gWhitegascontains=75%hydrocarbon=75100ghydrocarbon=0.75ghydrocarbon

Molar mass of C5 hydrocarbons (C5H12) is 72.148g/mol .

The enthalpy change of combustion (ΔHcombustion°) is 803.3kJ/mol .

The amount of fuel value of C5 hydrocarbons is calculated as,

Fuelvalue=ΔHcombustion°MFuelvalue=803.3kJ/mol72.148g/molFuelvalue=48.997kJ/g

The energy obtained by the 0.25g C5 hydrocarbon is calculated as,

Enrgyobtainedby1g=48.997kJEnrgyobtainedby0.25g=48.997kJ×0.25Enrgyobtainedby0.25g=12.249kJ

The calculated amount of fuel value of C6 hydrocarbon is 48.31kJ/g .

The energy obtained by the 0.75g of C6 hydrocarbon is calculated as,

Enrgyobtainedby1g=48.31kJEnrgyobtainedby0.75g=48.31kJ×0.75Enrgyobtainedby0.75g=36.23kJ

Therefore, the total amount of heat released when 1.00g white gas undergoes combustion is (12.249kJ)+(36.23kJ)=48.479kJ .

From all these calculations the amount of the compound C6H14 to heat 1.00kg of water from 25.0°C to 85.0°C is calculated as,

48.479kJheatobtained=1.00gmixtureofhydrocarbon1kJheatobtained=1.00gmixtureofhydrocarbon48.479kJ250.70kJheatobtained=1.00gC6H1448.479kJ×250.70kJ250.70kJheatobtained=5.17g_

Conclusion

  1. a) The amount of fuel value for the compound C6H14 is 48.31kJ/g_ .
  2. b) The amount of heat released during the combustion of 1.00kg of C6H14 is 48.31×103kJ_
  3. c) The amount of the compound C6H14 to heat 1.00kg of water from 25.0°C to 85.0°C is 5.189g_ .
  4. d) The amount of the white gas (25%C5+75%C6) to heat 1.00kg of water from 25.0°C to 85.0°C is 5.17g_

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Chapter 5 Solutions

Chemistry: The Science in Context (Fourth Edition)

Ch. 5.5 - Prob. 11PECh. 5.6 - Prob. 12PECh. 5.6 - Prob. 13PECh. 5.7 - Prob. 14PECh. 5.7 - Prob. 15PECh. 5.7 - Prob. 16PECh. 5.8 - Prob. 17PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143AP
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