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Energetics of Reactions - Fuels, Fire Triangle, and Ideal Fuels

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fuel is a substance that undergoes combustion to release energy, primarily in the form of heat and light. Common fuels include wood, coal, petroleum, and natural gas.

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Combustion is an exothermic chemical reaction between a fuel and an oxidant (usually O2O_2). The general equation for the complete combustion of a hydrocarbon is: CxHy+(x+y4)O2→xCO2+y2H2O+EnergyC_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + Energy.

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The Fire Triangle represents the three essential components required for a fire: Fuel (the combustible material), Heat (to reach the ignition temperature), and Oxygen (the oxidizing agent). Removing any one of these elements will extinguish the fire.

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Exothermic reactions are those where energy is released into the surroundings. In terms of enthalpy, the enthalpy of products is less than the enthalpy of reactants, meaning ΔH<0\Delta H < 0.

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An Ideal Fuel should have a high calorific value, a moderate ignition temperature (neither too high nor too low), a low rate of residue/ash production, and should be easy to transport and store.

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Calorific Value (CVCV) is the total heat energy released when a unit mass of fuel is completely burned in oxygen. It is measured in kJ/kgkJ/kg or J/gJ/g.

📐Formulae

Q=m×c×ΔTQ = m \times c \times \Delta T

Calorific Value=Heat Produced (Q)Mass of Fuel (mfuel)Calorific\ Value = \frac{Heat\ Produced\ (Q)}{Mass\ of\ Fuel\ (m_{fuel})}

ΔH=Hproducts−Hreactants\Delta H = H_{products} - H_{reactants}

Efficiency=Useful Energy OutputTotal Energy Input×100%Efficiency = \frac{Useful\ Energy\ Output}{Total\ Energy\ Input} \times 100\%

💡Examples

Problem 1:

During an experiment, 0.02 kg0.02\text{ kg} of a fuel is burned to heat 0.5 kg0.5\text{ kg} of water. The temperature of the water rises from 25∘C25^{\circ}C to 45∘C45^{\circ}C. If the specific heat capacity of water is 4.18 kJ/kg∘C4.18\text{ kJ/kg}^{\circ}C, calculate the heat energy absorbed by the water.

Solution:

Q=m×c×ΔTQ = m \times c \times \Delta T Q=0.5×4.18×(45−25)Q = 0.5 \times 4.18 \times (45 - 25) Q=0.5×4.18×20Q = 0.5 \times 4.18 \times 20 Q=41.8 kJQ = 41.8\text{ kJ}

Explanation:

The heat energy released by the fuel is transferred to the water. We use the mass of water (m=0.5 kgm = 0.5\text{ kg}), the specific heat capacity (c=4.18 kJ/kg∘Cc = 4.18\text{ kJ/kg}^{\circ}C), and the change in temperature (ΔT=20∘C\Delta T = 20^{\circ}C) to find the energy (QQ).

Problem 2:

Calculate the calorific value of a fuel if 2.5 kg2.5\text{ kg} of the fuel produces 75,000 kJ75,000\text{ kJ} of heat energy upon complete combustion.

Solution:

Calorific Value=Total Heat EnergyMass of FuelCalorific\ Value = \frac{Total\ Heat\ Energy}{Mass\ of\ Fuel} Calorific Value=75000 kJ2.5 kgCalorific\ Value = \frac{75000\text{ kJ}}{2.5\text{ kg}} Calorific Value=30,000 kJ/kgCalorific\ Value = 30,000\text{ kJ/kg}

Explanation:

Calorific value is the energy per unit mass. By dividing the total energy (75,000 kJ75,000\text{ kJ}) by the mass of the fuel consumed (2.5 kg2.5\text{ kg}), we determine the efficiency of the fuel in kJ/kgkJ/kg.

Problem 3:

A specific combustion reaction has a total reactant enthalpy of 2500 kJ2500\text{ kJ} and a total product enthalpy of 1800 kJ1800\text{ kJ}. Determine the enthalpy change (ΔH\Delta H) and state if the reaction is exothermic or endothermic.

Solution:

ΔH=Hproducts−Hreactants\Delta H = H_{products} - H_{reactants} ΔH=1800−2500\Delta H = 1800 - 2500 ΔH=−700 kJ\Delta H = -700\text{ kJ}

Explanation:

Since the enthalpy change ΔH\Delta H is negative (−700 kJ-700\text{ kJ}), the reaction releases energy to the surroundings and is therefore exothermic. This is typical for combustion of fuels.