Fuels and heat of combustion $E = Hm$

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 5-1Core

   At a petrol station, $2.0\,\text{kg}$ of petrol was bought for a chainsaw. The heat of combustion of petrol is $43\,\tfrac{\text{MJ}}{\text{kg}}$.

   How much energy is released when the petrol burns? Give the answer in megajoules.

   MJCheck
   Answer

   $86\,\text{MJ}$

   Show the solution

   Given

   petrol, mass $m = 2.0\,\text{kg}$ and heat of combustion $H = 43\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the energy $E$ released in burning, in megajoules

   Principle

   The chemical energy of petrol is released when it burns. The heat of combustion gives the energy released per kilogram, so the energy is the product of the heat of combustion and the mass.

   Formula

   $E = Hm$

   Conversions

   Not needed: the mass is already in kilograms and the answer is wanted in megajoules, like the heat of combustion.

   Substitution

   $E$$= 43\,\tfrac{\text{MJ}}{\text{kg}} \cdot 2.0\,\text{kg}$

   $= 86\,\text{MJ}$

   Answer

   $86\,\text{MJ}$

   Two significant figures like the mass $2.0\,\text{kg}$.

   Reasonableness

   One kilogram gives $43\,\text{MJ}$, so two kilograms give twice as much, about $2 \cdot 40\,\text{MJ} = 80\,\text{MJ}$. The result is of the right order of magnitude and positive.

2. Exercise 5-2Core

   On a winter day, $2.0\,\text{kg}$ of natural gas is burnt in a gas boiler. Read the heat of combustion of natural gas from the bar chart in the figure.

   How much energy is released from the gas? Give the answer in megajoules.

   MJCheck
   Answer

   $96\,\text{MJ}$

   Show the solution

   Given

   natural gas, mass $m = 2.0\,\text{kg}$; heat of combustion $H = 48\,\tfrac{\text{MJ}}{\text{kg}}$ is read from the figure

   Find

   the energy $E$ released, in megajoules

   Principle

   When the gas burns, its chemical energy is transformed into heat. The energy released is proportional to the mass, so we use the definition of the heat of combustion.

   Reading the graph

   The top bar of the bar chart is natural gas, and the number $48$ is printed at its end. The unit on the horizontal axis is $\text{MJ/kg}$, so $H = 48\,\tfrac{\text{MJ}}{\text{kg}}$; the end of the bar is also a little before the grid line $50$.

   Formula

   $E = Hm$

   Conversions

   Not needed: the mass is in kilograms and the heat of combustion in megajoules per kilogram, so the energy comes out in megajoules.

   Substitution

   $E$$= 48\,\tfrac{\text{MJ}}{\text{kg}} \cdot 2.0\,\text{kg}$

   $= 96\,\text{MJ}$

   Answer

   $96\,\text{MJ}$

   Two significant figures like the mass $2.0\,\text{kg}$.

   Reasonableness

   The heat of combustion of gas is about $50\,\tfrac{\text{MJ}}{\text{kg}}$, so two kilograms give about $100\,\text{MJ}$. The result is slightly smaller, because $48 < 50$.

3. Exercise 5-3

   A coal-fired power plant burns $12\,000\,\text{kg}$ of coal per hour, with a heat of combustion of $26\,\tfrac{\text{MJ}}{\text{kg}}$.

   How much energy is released from the coal in an hour? Give the answer in megajoules.

   MJCheck
   Answer

   $310\,000\,\text{MJ}$

   Show the solution

   Given

   coal, mass $m = 12\,000\,\text{kg}$ and heat of combustion $H = 26\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the energy $E$ released in one hour, in megajoules

   Principle

   The chemical energy of coal is released when it burns, and the energy released is the product of the heat of combustion and the mass. The mass is already in kilograms, as the heat of combustion requires.

   Formula

   $E = Hm$

   Conversions

   Not needed: the mass is in kilograms and the answer is wanted in megajoules, like the heat of combustion.

   Substitution

   $E$$= 26\,\tfrac{\text{MJ}}{\text{kg}} \cdot 12\,000\,\text{kg}$

   $= 312\,000\,\text{MJ}$

   Answer

   $310\,000\,\text{MJ}$

   Two significant figures like the heat of combustion $26\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   Roughly

   $10\,000\,\text{kg} \cdot 30\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 300\,000\,\text{MJ}$

   The energy for one hour at a power plant is large, because about ten tonnes of fuel are used.

4. Exercise 5-4

   $400\,\text{kg}$ of milled peat with a heat of combustion of $10\,\tfrac{\text{MJ}}{\text{kg}}$ is taken from the storage of a heating plant to be burnt.

   How much energy is released from the peat? Give the answer in megajoules.

   MJCheck
   Answer

   $4000\,\text{MJ}$

   Show the solution

   Given

   milled peat, mass $m = 400\,\text{kg}$ and heat of combustion $H = 10\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the energy $E$ released, in megajoules

   Principle

   When peat burns, its chemical energy is released, and the energy is the product of the heat of combustion and the mass.

   Formula

   $E = Hm$

   Conversions

   Not needed: the mass is already in kilograms.

   Substitution

   $E$$= 10\,\tfrac{\text{MJ}}{\text{kg}} \cdot 400\,\text{kg}$

   $= 4000\,\text{MJ}$

   Answer

   $4000\,\text{MJ}$

   Or $4.0 \cdot 10^3\,\text{MJ}$, two significant figures like the heat of combustion $10\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   The heat of combustion of peat is small, so a hundred kilograms gives only $1000\,\text{MJ}$. Four times the mass gives four times the energy, $4000\,\text{MJ}$.

5. Exercise 5-5

   The energy content of a hiker's nut mix is $24\,\tfrac{\text{MJ}}{\text{kg}}$ (example value). A handful of nuts weighs $0.040\,\text{kg}$.

   How much energy do you get from the handful? Give the answer in megajoules.

   MJCheck
   Answer

   $0.96\,\text{MJ}$

   Show the solution

   Given

   nuts, energy content $e = 24\,\tfrac{\text{MJ}}{\text{kg}}$ and mass $m = 0.040\,\text{kg}$

   Find

   the energy $E$ contained in the nuts, in megajoules

   Principle

   The energy of food is calculated in the same way as for a fuel: the energy content is the energy the body can use per kilogram, so the energy is the product of it and the mass. The energy content is not a heat of combustion, even though the unit is the same.

   Formula

   $E = em$

   Conversions

   Not needed: the mass is in kilograms and the answer is wanted in megajoules.

   Substitution

   $E$$= em$

   $= 24\,\tfrac{\text{MJ}}{\text{kg}} \cdot 0.040\,\text{kg}$

   $= 0.96\,\text{MJ}$

   Answer

   $0.96\,\text{MJ}$

   Two significant figures like the mass $0.040\,\text{kg}$.

   Reasonableness

   An estimate with rounded numbers: $25\,\tfrac{\text{MJ}}{\text{kg}} \cdot 0.04\,\text{kg} = 1\,\text{MJ}$, so $0.96\,\text{MJ}$ is of the right order of magnitude. It is about $1000\,\text{kJ}$, the typical energy of a small snack.

6. Exercise 5-6

   The heat of combustion of petrol is $43\,\tfrac{\text{MJ}}{\text{kg}}$. What does this mean?

   Choose an option.

   • Petrol burns with a hotter flame than other fuels, and the number $43\,\text{MJ}$ tells the temperature of the flame.
   • To obtain one megajoule of energy, $43\,\text{kg}$ of petrol always has to be burned, regardless of the amount.
   • Every kilogram of petrol burned creates $43\,\text{MJ}$ of new energy that did not exist before burning.
   • When one kilogram of petrol burns, an energy of $43\,\text{MJ}$ is released.

   Answer
   I have done my solution.
   Answer

   When one kilogram of petrol burns, an energy of $43\,\text{MJ}$ is released.

   Show the solution

   Find

   what the heat of combustion of petrol $H = 43\,\tfrac{\text{MJ}}{\text{kg}}$ tells

   Principle

   The heat of combustion is the energy released per unit of mass when a fuel burns, so its unit $\text{MJ/kg}$ is energy divided by mass. It is not a temperature, and energy is not created from nothing: chemical energy is transformed into heat and light.

   Model answer

   When one kilogram of petrol burns, an energy of $43\,\text{MJ}$ is released. Two kilograms release twice as much, $E = Hm = 43\,\tfrac{\text{MJ}}{\text{kg}} \cdot 2.0\,\text{kg} = 86\,\text{MJ}$. The heat of combustion does not tell the temperature of the flame, and it does not mean that energy would be created anew.

7. Exercise 5-7

   In a pellet burner, an energy of $255\,\text{MJ}$ is to be released from wood pellets. The heat of combustion of the pellets is $17\,\tfrac{\text{MJ}}{\text{kg}}$.

   How many kilograms of pellets must be burned? Give the answer in kilograms.

   kgCheck
   Answer

   $15\,\text{kg}$

   Show the solution

   Given

   the energy to be released is $E = 255\,\text{MJ}$ and the heat of combustion of the pellets is $H = 17\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the mass $m$ of pellets to be burned, in kilograms

   Principle

   The chemical energy of the pellets is released when they burn, and the energy released is the product of the heat of combustion and the mass. The mass is obtained by dividing the energy needed by the heat of combustion.

   Formula

   $E = Hm$, so $m = \dfrac{E}{H}$

   Conversions

   Not needed: the energy is in megajoules and the heat of combustion in megajoules per kilogram, so the mass comes out in kilograms.

   Substitution

   $m$$= \dfrac{255\,\text{MJ}}{17\,\tfrac{\text{MJ}}{\text{kg}}}$

   $= 15\,\text{kg}$

   Answer

   $15\,\text{kg}$

   Two significant figures like the heat of combustion $17\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   One kilogram gives $17\,\text{MJ}$, so $15\,\text{kg}$ gives $15 \cdot 17\,\text{MJ} = 255\,\text{MJ}$, which is the energy in the problem.

8. Exercise 5-8

   When three kilograms of natural gas burn, an energy of $144\,\text{MJ}$ is released. How much is this in kilowatt-hours? Use the relation $1\,\text{kWh} = 3.6\,\text{MJ}$. Give the answer in kilowatt-hours.

   kWhCheck
   Answer

   $40\,\text{kWh}$

   Show the solution

   Given

   mass of natural gas $m = 3.0\,\text{kg}$ and energy released $E = 144\,\text{MJ}$

   Find

   the energy $E$ in kilowatt-hours

   Principle

   Megajoules and kilowatt-hours are units of the same quantity, energy. The conversion is done by substituting the unit relation $1\,\text{kWh} = 3.6\,\text{MJ}$, so the value of the energy does not change, only its unit.

   Formula

   $E$ in kilowatt-hours:

   $E$$= \dfrac{144\,\text{MJ}}{3.6\,\text{MJ/kWh}}$

   Conversions

   $1\,\text{kWh} = 3.6\,\text{MJ}$, so $1\,\text{MJ} = 1\,\text{kWh} : 3.6$.

   Substitution

   $E$$= \dfrac{144\,\text{MJ}}{3.6\,\text{MJ/kWh}}$

   $= 40\,\text{kWh}$

   Answer

   $40\,\text{kWh}$

   Two significant figures like the energy $144\,\text{MJ}$ (the conversion factor $3.6$ is exact).

   Reasonableness

   One kilowatt-hour is $3.6\,\text{MJ}$, so $40\,\text{kWh}$ is $40 \cdot 3.6\,\text{MJ} = 144\,\text{MJ}$, as in the problem. The number of kilowatt-hours is smaller than the number of megajoules, because one kilowatt-hour is a larger amount of energy than one megajoule.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 5-9Core

   The nutritional label on a loaf of bread (example) says: energy $1200\,\text{kJ}$ per hundred grams. A slice of bread weighs $40\,\text{g}$.

   How much energy is in the slice? Give the answer in kilojoules.

   kJCheck
   Answer

   $480\,\text{kJ}$

   Show the solution

   Given

   according to the label $1200\,\text{kJ}$ per hundred grams, mass of the slice $m = 40\,\text{g}$

   Find

   the energy $E$ of the slice, in kilojoules

   Principle

   The label gives the energy content of the food, that is the energy the body can use, per unit of mass. It is not a heat of combustion, but the calculation is done in the same way: the label is converted to megajoules per kilogram and multiplied by the mass.

   Formula

   $E = em$, where the energy content $e$ is the energy on the label divided by the mass on the label

   Conversions

   $1200\,\text{kJ}$$= 1200 \cdot (1 : 1000)\,\text{MJ}$

   $= 1.2\,\text{MJ}$

   $100\,\text{g}$$= 100 \cdot (1 : 1000)\,\text{kg}$

   $= 0.100\,\text{kg}$

   $40\,\text{g}$$= 40 \cdot (1 : 1000)\,\text{kg}$

   $= 0.040\,\text{kg}$

   Substitution

   $e$$= 1.2\,\text{MJ} : 0.100\,\text{kg}$

   $= 12\,\tfrac{\text{MJ}}{\text{kg}}$

   $E$$= 12\,\tfrac{\text{MJ}}{\text{kg}} \cdot 0.040\,\text{kg}$

   $= 0.48\,\text{MJ}$

   $= 480\,\text{kJ}$

   Answer

   $480\,\text{kJ}$

   Two significant figures like the mass $40\,\text{g}$.

   Reasonableness

   The slice is $40\,\%$ of a hundred grams, and $40\,\%$ of $1200\,\text{kJ}$ is $0.4 \cdot 1200\,\text{kJ} = 480\,\text{kJ}$. The result is smaller than $1200\,\text{kJ}$, as it should be.

2. Exercise 5-10Core

   A diesel generator on a building site must release $1.8\,\text{GJ}$ of energy from fuel. The heat of combustion of diesel is $43\,\tfrac{\text{MJ}}{\text{kg}}$.

   How many kilograms of diesel must be burnt? Give the answer in kilograms.

   kgCheck
   Answer

   $42\,\text{kg}$

   Show the solution

   Given

   the energy released must be $E = 1.8\,\text{GJ}$, heat of combustion of diesel $H = 43\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the mass $m$ of diesel to be burnt, in kilograms

   Principle

   By the law $E = Hm$ the energy is the product of the heat of combustion and the mass. Now the energy is known and the mass is wanted, so the law is solved for the mass.

   Formula

   $E = Hm$, so $m = \dfrac{E}{H}$

   Conversions

   $1.8\,\text{GJ}$$= 1.8 \cdot 1000\,\text{MJ}$

   $= 1800\,\text{MJ}$

   Substitution

   $m$$= 1800\,\text{MJ} : 43\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 41.86\,\text{kg}$

   Answer

   $42\,\text{kg}$

   Two significant figures like the data $1.8\,\text{GJ}$ and $43\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   One kilogram gives about $43\,\text{MJ}$, so $42\,\text{kg}$ gives about $42 \cdot 43\,\text{MJ} \approx 1800\,\text{MJ}$, which is $1.8\,\text{GJ}$.

3. Exercise 5-11

   In the pellet burner of a detached house, $40\,\text{kg}$ of wood pellets with a heat of combustion of $17\,\tfrac{\text{MJ}}{\text{kg}}$ are burnt. The efficiency of the boiler is $85\,\%$.

   How much useful energy is obtained from the pellets to heat the house? Give the answer in megajoules.

   MJCheck
   Answer

   $580\,\text{MJ}$

   Show the solution

   Given

   wood pellets, mass $m = 40\,\text{kg}$ and heat of combustion $H = 17\,\tfrac{\text{MJ}}{\text{kg}}$; efficiency of the boiler $\eta = 85\,\% = 0.85$

   Find

   the useful energy $E_\text{useful}$ in megajoules

   Principle

   The pellets release the energy $E_\text{in} = Hm$. Efficiency is the share of the energy taken in that is obtained as useful energy (chapter Efficiency), so only part of the released energy ends up heating the house.

   Formula

   $E_\text{in}$$= Hm$

   $\eta$$= \dfrac{E_\text{useful}}{E_\text{in}}$

   so $E_\text{useful} = \eta\, E_\text{in}$

   Conversions

   Not needed: the mass is in kilograms and $85\,\% = 0.85$.

   Substitution

   $E_\text{in}$$= 17\,\tfrac{\text{MJ}}{\text{kg}} \cdot 40\,\text{kg}$

   $= 680\,\text{MJ}$

   $E_\text{useful}$$= 0.85 \cdot 680\,\text{MJ}$

   $= 578\,\text{MJ}$

   Answer

   $580\,\text{MJ}$

   Two significant figures like the data $40\,\text{kg}$ and $85\,\%$.

   Reasonableness

   The useful energy is smaller than the $680\,\text{MJ}$ released, because part of the heat leaves with the flue gases. Estimate: $0.85 \approx 0.9$ and $0.9 \cdot 680\,\text{MJ} \approx 610\,\text{MJ}$, so $578\,\text{MJ}$ is of the right order of magnitude.

4. Exercise 5-12

   A passenger car needs $48\,\text{MJ}$ of useful energy to drive $100\,\text{km}$. The efficiency of the engine is $25\,\%$. The heat of combustion of petrol is $43\,\tfrac{\text{MJ}}{\text{kg}}$.

   How many kilograms of petrol are used on a $100\,\text{km}$ journey? Give the answer in kilograms.

   kgCheck
   Answer

   $4.5\,\text{kg}$

   Show the solution

   Given

   useful energy $E_\text{useful} = 48\,\text{MJ}$, efficiency $\eta = 25\,\% = 0.25$, heat of combustion of petrol $H = 43\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   the mass $m$ of petrol, in kilograms

   Principle

   The engine transforms only part of the energy of the petrol into motion (chapter Efficiency). First we use the efficiency to find the energy needed from the petrol, and then the law $E = Hm$ to find the mass.

   Formula

   $\eta$$= \dfrac{E_\text{useful}}{E_\text{in}}$

   so

   $E_\text{in}$$= \dfrac{E_\text{useful}}{\eta}$

   $E_\text{in} = Hm$, so $m = \dfrac{E_\text{in}}{H}$

   Conversions

   Not needed: $25\,\% = 0.25$ and the energies are in megajoules.

   Substitution

   $E_\text{in}$$= 48\,\text{MJ} : 0.25$

   $= 192\,\text{MJ}$

   $m$$= 192\,\text{MJ} : 43\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 4.465\,\text{kg}$

   Answer

   $4.5\,\text{kg}$

   Two significant figures like the data $48\,\text{MJ}$ and $25\,\%$.

   Reasonableness

   The density of petrol is about $0.74\,\tfrac{\text{kg}}{\textsf{l}}$, so $4.5\,\text{kg}$ is about $6\,\textsf{l}$, a typical consumption per hundred kilometres. The energy of the petrol, $192\,\text{MJ}$, is four times the useful energy, because the efficiency is one quarter.

5. Exercise 5-13

   The table gives example data on the effective heat of combustion of firewood at different moisture contents. Heating a sauna requires an energy of $200\,\text{MJ}$ released from firewood.

   Using the heats of combustion in the table, calculate the mass of wood needed when the moisture content is $20\,\%$ and when it is $40\,\%$. How many kilograms more moist ($40\,\%$) wood is needed than air-dried ($20\,\%$) wood? Give the answer in kilograms.

   Example data: effective heat of combustion of firewood by moisture content

   Moisture content $(\%)$	$H\,(\text{MJ/kg})$
   0	19.0
   10	16.9
   20	14.7
   30	12.6
   40	10.4
   50	8.3

   kgCheck
   Answer

   $5.6\,\text{kg}$

   Show the solution

   Given

   the energy needed $E = 200\,\text{MJ}$; from the table the heat of combustion $H_{20} = 14.7\,\tfrac{\text{MJ}}{\text{kg}}$ at moisture content $20\,\%$ and $H_{40} = 10.4\,\tfrac{\text{MJ}}{\text{kg}}$ at $40\,\%$

   Find

   the difference of the masses $\Delta m = m_{40} - m_{20}$ in kilograms

   Principle

   The same energy has to be released from both kinds of wood. By the law $E = Hm$, the mass is the energy divided by the heat of combustion. Moist wood gives less energy per kilogram, because part of its mass is water, which does not burn and whose evaporation consumes energy.

   Formula

   $E = Hm$, so $m = \dfrac{E}{H}$, and $\Delta m = m_{40} - m_{20}$

   Conversions

   Not needed: the energy is in megajoules and the heats of combustion in megajoules per kilogram.

   Substitution

   $m_{20}$$= 200\,\text{MJ} : 14.7\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 13.61\,\text{kg}$

   $m_{40}$$= 200\,\text{MJ} : 10.4\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 19.23\,\text{kg}$

   The difference is

   $\Delta m$$= 19.23\,\text{kg} - 13.61\,\text{kg}$

   $= 5.62\,\text{kg}$

   Answer

   $5.6\,\text{kg}$

   More moist wood is needed, two significant figures like the data $200\,\text{MJ}$.

   Reasonableness

   The ratio of the heats of combustion is $14.7 : 10.4 = 1.41$, so the mass of the moist wood is about $1.41$ times as large: $1.41 \cdot 13.61\,\text{kg} = 19.2\,\text{kg}$. Moist wood gives less energy per kilogram, so more mass is needed, and the difference is positive.

6. Exercise 5-14

   A student solved a problem: “A power plant burns $3.0\,\text{t}$ of coal with a heat of combustion of $26\,\tfrac{\text{MJ}}{\text{kg}}$. The energy is $E = Hm = 26 \cdot 3.0\,\text{MJ} = 78\,\text{MJ}$.” Where is the student’s mistake?

   Choose an option.

   • The mass was not converted to kilograms: $3.0\,\text{t} = 3000\,\text{kg}$, so the correct result is $78\,000\,\text{MJ}$.
   • The result should have been given in kilowatt-hours, so the correct result is $78\,\text{kWh}$.
   • The heat of combustion of coal is wrong, because the correct value would be $43\,\tfrac{\text{MJ}}{\text{kg}}$, so the result is $129\,\text{MJ}$.
   • The heat of combustion should have been divided by the mass, so the correct result is $26 : 3.0 = 8.7\,\text{MJ}$.

   Answer
   I have done my solution.
   Answer

   The mass was not converted to kilograms: $3.0\,\text{t} = 3000\,\text{kg}$, so the correct result is $78\,000\,\text{MJ}$.

   Show the solution

   Find

   where the student’s calculation goes wrong and what the correct result is

   Principle

   The heat of combustion is energy per kilogram, so in the formula $E = Hm$ the mass must be substituted in kilograms. A tonne is not a kilogram: $1\,\text{t} = 1000\,\text{kg}$.

   Model answer

   The student substituted the mass in tonnes, although the heat of combustion is in the unit $\text{MJ/kg}$. Correctly, $3.0\,\text{t} = 3.0 \cdot 1000\,\text{kg} = 3000\,\text{kg}$ and $E = Hm = 26\,\tfrac{\text{MJ}}{\text{kg}} \cdot 3000\,\text{kg} = 78\,000\,\text{MJ}$. The student’s result $78\,\text{MJ}$ is a thousand times too small. The formula, the heat of combustion of coal and the unit $\text{MJ}$ of the answer were correct. The thousandfold error also shows in an estimate: one kilogram of coal already gives $26\,\text{MJ}$, so three tonnes give much more than $78\,\text{MJ}$.

7. Exercise 5-15

   In an example measurement, a pellet burner and its pellets stood on a scale, and the burner burned steadily the whole time. The reading $m$ of the scale was recorded every $10\,\text{min}$.

   Draw a graph of the mass as a function of time and determine from its slope how many kilograms of pellets burn per minute. The heat of combustion of the pellets is $17\,\tfrac{\text{MJ}}{\text{kg}}$. What is the power released in the burning? Give the answer in kilowatts.

   Example measurement: mass of the pellet burner and the pellets

   $t\,(\text{min})$	$m\,(\text{kg})$
   0	12.00
   10	11.34
   20	10.71
   30	10.04
   40	9.40

   kWCheck
   Answer

   $18\,\text{kW}$

   Show the solution

   Given

   a table $(t, m)$, heat of combustion of the pellets $H = 17\,\tfrac{\text{MJ}}{\text{kg}}$; the burner burns steadily

   Find

   the power $P$ released in the burning, in kilowatts

   Principle

   The burner burns steadily, so the mass decreases steadily and the graph $m(t)$ is a straight line; its slope tells the mass burned per minute. The burned mass releases an energy $E = Hm$, and power is energy per unit of time, $P = E : t$ (chapter Power).

   Reading the graph

   When the points $(t, m)$ are plotted, they lie roughly on the same descending line, because the mass decreases by about $0.65\,\text{kg}$ every $10$ minutes (the small scatter of the measured points is part of the measurement). A suitable line is drawn through the points and its slope is read from the ends of the line, which pass through the points $(0\,\text{min}, 12.00\,\text{kg})$ and $(40\,\text{min}, 9.40\,\text{kg})$:

   $k$$= \dfrac{9.40\,\text{kg} - 12.00\,\text{kg}}{40\,\text{min} - 0\,\text{min}}$

   $= -0.065\,\tfrac{\text{kg}}{\text{min}}$

   The minus sign shows that the mass decreases, so the rate at which mass burns is $0.065\,\tfrac{\text{kg}}{\text{min}}$.

   Formula

   $E = Hm$ and $P = \dfrac{E}{t}$, together $P = H \cdot \dfrac{m}{t}$

   Conversions

   $1\,\text{min} = 60\,\text{s}$, so $0.065\,\tfrac{\text{kg}}{\text{min}} = 0.065\,\text{kg} : 60\,\text{s}$. The heat of combustion is

   $17\,\tfrac{\text{MJ}}{\text{kg}}$$= 17 \cdot 10^{6}\,\tfrac{\text{J}}{\text{kg}}$

   and $1\,\text{W} = 1\,\tfrac{\text{J}}{\text{s}}$.

   Substitution

   $P$$= 17 \cdot 10^{6}\,\tfrac{\text{J}}{\text{kg}} \cdot \dfrac{0.065\,\text{kg}}{60\,\text{s}}$

   $= 18\,417\,\text{W}$

   $= 18.4\,\text{kW}$

   Answer

   $18\,\text{kW}$

   Two significant figures like the heat of combustion $17\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   In a minute $0.065\,\text{kg}$ burns, which releases $17\,\tfrac{\text{MJ}}{\text{kg}} \cdot 0.065\,\text{kg} = 1.1\,\text{MJ}$. That is

   $1.1\,\text{MJ} : 60\,\text{s}$$= 0.018\,\text{MJ/s}$

   $= 18\,\text{kW}$

   which suits the fuel power of the pellet boiler of a detached house. The power is the power released, not the useful power received by the house.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 5-16

   To heat a cottage, $120\,\text{MJ}$ of energy must be released from fuel. The energy can be obtained either from air-dried firewood ($15\,\tfrac{\text{MJ}}{\text{kg}}$) or from light heating oil ($43\,\tfrac{\text{MJ}}{\text{kg}}$). Calculate the mass of each fuel needed and say how many kilograms more wood than oil are needed. Also explain why more wood is needed. Give the answer in kilograms.

   kgCheck
   Answer

   $5.2\,\text{kg}$

   Show the solution

   Given

   the energy needed $E = 120\,\text{MJ}$, air-dried firewood $H_\text{wood} = 15\,\tfrac{\text{MJ}}{\text{kg}}$ and heating oil $H_\text{oil} = 43\,\tfrac{\text{MJ}}{\text{kg}}$

   Find

   how many kilograms more wood than oil are needed, $\Delta m = m_\text{wood} - m_\text{oil}$

   Principle

   The same energy $E$ must be released from both fuels. By the law $E = Hm$ the mass is the energy divided by the heat of combustion, so a smaller heat of combustion needs a larger mass.

   Formula

   $E = Hm$, so $m = \dfrac{E}{H}$ for each fuel, and $\Delta m = m_\text{wood} - m_\text{oil}$

   Conversions

   Not needed: the energy is in megajoules and the heats of combustion in megajoules per kilogram.

   Substitution

   $m_\text{wood}$$= 120\,\text{MJ} : 15\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 8.0\,\text{kg}$

   $m_\text{oil}$$= 120\,\text{MJ} : 43\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 2.79\,\text{kg}$

   Difference:

   $\Delta m$$= 8.0\,\text{kg} - 2.79\,\text{kg}$

   $= 5.21\,\text{kg}$

   Answer

   $5.2\,\text{kg}$

   More wood is needed, two significant figures like the data $120\,\text{MJ}$ and $15\,\tfrac{\text{MJ}}{\text{kg}}$. More wood is needed because one kilogram of wood contains less chemical energy than one kilogram of oil.

   Reasonableness

   The mass of wood is about $2.9$ times the mass of oil, since $43 : 15 = 2.87$. So $2.87 \cdot 2.79\,\text{kg} \approx 8.0\,\text{kg}$, and the difference is $2.87 - 1 = 1.87$ times the mass of oil: $1.87 \cdot 2.79\,\text{kg} \approx 5.2\,\text{kg}$, as it should be.

2. Exercise 5-17

   A hiker carries food with an energy content of $18\,\tfrac{\text{MJ}}{\text{kg}}$ (example value). There is $2.4\,\text{kg}$ of food, and the hiker needs $12\,\text{MJ}$ of energy from food per day.

   How many daily rations does the food provide? Give the answer as a decimal number and justify your calculation.

   daily rationsCheck
   Answer

   $3.6$

   Show the solution

   Given

   energy content of the food $e = 18\,\tfrac{\text{MJ}}{\text{kg}}$, mass $m = 2.4\,\text{kg}$, daily consumption $12\,\text{MJ}$

   Find

   the number of daily rations $N$ that the food provides (as a decimal number)

   Principle

   The total energy obtainable from the food is found by multiplying the energy content by the mass, as with a fuel by the law $E = Hm$. The number of rations is how many times the daily consumption fits into the total energy, so $N = E : \text{daily consumption}$.

   Formula

   $E = em$ and the number of days $N = \dfrac{E}{12\,\text{MJ}}$

   Conversions

   Not needed: the mass is in kilograms and the energy in megajoules.

   Substitution

   $E$$= 18\,\tfrac{\text{MJ}}{\text{kg}} \cdot 2.4\,\text{kg}$

   $= 43.2\,\text{MJ}$

   $N$$= 43.2\,\text{MJ} : 12\,\text{MJ}$

   $= 3.6$

   Answer

   $3.6$

   Daily rations, two significant figures like the data $2.4\,\text{kg}$ and $18\,\tfrac{\text{MJ}}{\text{kg}}$.

   Reasonableness

   The need for three rations is $36\,\text{MJ}$ and for four $48\,\text{MJ}$. There is $43.2\,\text{MJ}$ of food, so the result lies between three and four: three full days and part of a fourth.

3. Exercise 5-18

   A peat-fired power plant burns $36\,\text{t}$ of milled peat per hour, with a heat of combustion of $10\,\tfrac{\text{MJ}}{\text{kg}}$. The efficiency of the plant in electricity production is $35\,\%$.

   What is the electric power of the power plant? Give the answer in megawatts. Remember: $1\,\text{W} = 1\,\text{J/s}$ and one hour is $3600\,\text{s}$.

   MWCheck
   Answer

   $35\,\text{MW}$

   Show the solution

   Given

   peat $m = 36\,\text{t}$ per hour, $H = 10\,\tfrac{\text{MJ}}{\text{kg}}$, efficiency of electricity production $\eta = 35\,\% = 0.35$, time $t = 1\,\text{h}$

   Find

   the electric power $P$ in megawatts

   Principle

   In one hour the peat releases the energy $E_\text{in} = Hm$. By the efficiency, only the share $\eta$ of it becomes electricity (chapter Efficiency). Power is energy per unit of time, so power = energy : time (chapter Power P = ΔE/Δt).

   Formula

   $E_\text{in} = Hm$, $E_\text{useful} = \eta\, E_\text{in}$ and $P = \dfrac{E_\text{useful}}{t}$, together $P = \dfrac{\eta H m}{t}$

   Conversions

   $36\,\text{t}$$= 36 \cdot 1000\,\text{kg}$

   $= 36\,000\,\text{kg}$

   $1\,\text{h} = 3600\,\text{s}$. Since $1\,\text{MJ/s} = 1\,\text{MW}$, the unit of power comes out directly in megawatts.

   Substitution

   $E_\text{in}$$= 10\,\tfrac{\text{MJ}}{\text{kg}} \cdot 36\,000\,\text{kg}$

   $= 360\,000\,\text{MJ}$

   $E_\text{useful}$$= 0.35 \cdot 360\,000\,\text{MJ}$

   $= 126\,000\,\text{MJ}$

   $P$$= 126\,000\,\text{MJ} : 3600\,\text{s}$

   $= 35\,\text{MJ/s}$

   $= 35\,\text{MW}$

   Answer

   $35\,\text{MW}$

   Two significant figures like the data $10\,\tfrac{\text{MJ}}{\text{kg}}$ and $35\,\%$.

   Reasonableness

   If all the energy of the peat became electricity, the power would be $360\,000\,\text{MJ} : 3600\,\text{s} = 100\,\text{MW}$. With an efficiency of $35\,\%$ we get $35\,\%$ of that, $35\,\text{MW}$, which is a typical power for a small power plant.

4. Exercise 5-19

   The table gives example data on the useful energy needed each month to heat a detached house. The house can be heated either with a pellet boiler (heat of combustion of pellets $17\,\tfrac{\text{MJ}}{\text{kg}}$, efficiency $85\,\%$) or with light heating oil (heat of combustion $43\,\tfrac{\text{MJ}}{\text{kg}}$, efficiency $90\,\%$).

   a) Calculate the heating energy of the year in kilowatt-hours and in megajoules.

   b) Calculate the mass of pellets burned in the pellet boiler in a year.

   c) Calculate the mass of oil used in a year and compare it with the mass of pellets. Explain why the ratio of the masses is not the same as the ratio of the heats of combustion.

   d) Because of poor adjustment, the efficiency of the pellet boiler drops to $70\,\%$. How many kilograms more pellets are then used in a year, and why?

   Enter the result of part b) in kilograms in the answer box.

   Example data: useful energy needed to heat the house each month

   Month	$E\,(\text{kWh})$
   January	2100
   February	1900
   March	1600
   April	1100
   May	600
   June	250
   July	200
   August	250
   September	600
   October	1100
   November	1600
   December	2000

   kgCheck
   Answer

   $3300\,\text{kg}$

   Show the solution

   Given

   the monthly energies in the table (sum below); pellets $H_\text{p} = 17\,\tfrac{\text{MJ}}{\text{kg}}$, $\eta_\text{p} = 85\,\% = 0.85$; oil $H_\text{oil} = 43\,\tfrac{\text{MJ}}{\text{kg}}$, $\eta_\text{oil} = 90\,\% = 0.90$; efficiency with poor adjustment $\eta_\text{p}' = 0.70$

   Find

   the yearly consumption of pellets $m_\text{p}$ in kilograms; also the useful energy of the year $E_\text{useful}$, the mass of oil $m_\text{oil}$ and the extra consumption of pellets at efficiency $0.70$

   Principle

   The energy in the table is the useful energy that ends up in the heating. The efficiency is the share of the energy released from the fuel that becomes useful energy, so the energy released is $E_\text{in} = E_\text{useful} : \eta$ and the mass of the fuel follows from the law $E = Hm$. Energy is conserved: what does not end up in the heating is lost in the flue gases and the boiler (chapter Efficiency).

   Formula

   $E_\text{in}$$= \dfrac{E_\text{useful}}{\eta}$

   $E_\text{in} = Hm$, so $m = \dfrac{E_\text{useful}}{\eta H}$

   Conversions

   $1\,\text{kWh} = 3.6\,\text{MJ}$, so $E_\text{useful} = 13\,300 \cdot 3.6\,\text{MJ}$. Efficiencies $85\,\% = 0.85$, $90\,\% = 0.90$ and $70\,\% = 0.70$.

   Substitution

   The sum of the months is

   $2100 + 1900 + 1600 + 1100 + 600 + 250 + 200 + 250 + 600 + 1100 + 1600 + 2000$

   $= 13\,300$

   so

   $E_\text{useful}$$= 13\,300\,\text{kWh}$

   $= 13\,300 \cdot 3.6\,\text{MJ}$

   $= 47\,880\,\text{MJ}$

   With pellets

   $E_\text{in}$$= 47\,880\,\text{MJ} : 0.85$

   $= 56\,329\,\text{MJ}$

   $m_\text{p}$$= 56\,329\,\text{MJ} : 17\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 3313\,\text{kg}$

   With oil

   $E_\text{in}$$= 47\,880\,\text{MJ} : 0.90$

   $= 53\,200\,\text{MJ}$

   $m_\text{oil}$$= 53\,200\,\text{MJ} : 43\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 1237\,\text{kg}$

   At efficiency $0.70$ the pellets need

   $47\,880\,\text{MJ} : 0.70$

   $= 68\,400\,\text{MJ}$

   $m_\text{p}'$

   $= 68\,400\,\text{MJ} : 17\,\tfrac{\text{MJ}}{\text{kg}}$

   $= 4024\,\text{kg}$

   so the extra consumption is $4024\,\text{kg} - 3313\,\text{kg} = 710\,\text{kg}$.

   Answer

   The heating energy of the year is $13\,300\,\text{kWh} = 47\,900\,\text{MJ}$. The pellets used are $3300\,\text{kg}$, about $3.3$ tonnes, two significant figures like the efficiency $85\,\%$ and the heat of combustion $17\,\tfrac{\text{MJ}}{\text{kg}}$. The oil used is $1200\,\text{kg}$, and with poor adjustment about $710\,\text{kg}$ more pellets are used. The mass of pellets is about $2.7$ times the mass of oil, but the ratio of the heats of combustion is $43 : 17 = 2.5$. The difference is due to the efficiencies: the efficiency of the oil boiler is higher ($0.90$ instead of $0.85$), so even less oil is needed, and $2.5 \cdot (0.90 : 0.85) = 2.7$. A drop in efficiency increases the consumption, because the same useful energy has to be obtained from a larger energy released; the losses grow.

   Reasonableness

   A rough estimate: $13\,300\,\text{kWh}$ is about $48\,000\,\text{MJ}$, and pellets give $17\,\text{MJ}$ per kilogram, so without losses $48\,000 : 17 = 2820\,\text{kg}$ would be needed. Because of the losses the amount is larger, $2820\,\text{kg} : 0.85 = 3320\,\text{kg}$, which agrees with the result. Consumption is highest in the winter months, and in summer heating energy is needed only for hot water.

5. Exercise 5-20

   A housing company needs $36\,000\,\text{MJ}$ of useful energy per year for heating (example data). The table gives three boiler options. The fuel is assumed to burn completely and the efficiency of the boiler to stay constant.

   a) For each option, calculate the mass of fuel needed in a year.

   b) How many kilograms more firewood than pellets are used in a year? Use the unrounded masses from part a).

   c) As the pellet boiler ages, its efficiency drops to $75\,\%$. Calculate the consumption of pellets then and explain how the efficiency affects the amount of fuel needed.

   d) Justify why the mass of oil used is clearly smaller than the mass of wood, even though the efficiencies of the boilers are of the same order of magnitude.

   Enter the result of part b) in kilograms in the answer box.

   Example data: three boiler options

   Boiler	$H\,(\text{MJ/kg})$	$\eta\,(\%)$
   Wood boiler (air-dried wood)	15	70
   Pellet boiler	17	85
   Oil boiler	43	90

   kgCheck
   Answer

   $940\,\text{kg}$

   Show the solution

   Given

   the useful energy of the year $E_\text{useful} = 36\,000\,\text{MJ}$; from the table wood $H = 15\,\tfrac{\text{MJ}}{\text{kg}}$, $\eta = 0.70$; pellets $H = 17\,\tfrac{\text{MJ}}{\text{kg}}$, $\eta = 0.85$; oil $H = 43\,\tfrac{\text{MJ}}{\text{kg}}$, $\eta = 0.90$

   Find

   the difference of the masses of wood and pellets $\Delta m = m_\text{wood} - m_\text{pellet}$ in kilograms; also the mass of each fuel and the consumption of pellets at efficiency $0.75$

   Principle

   A model: the energy released from the fuel is $E_\text{in} = Hm$, and only the share $\eta$ of it becomes useful energy, $E_\text{useful} = \eta H m$. The smaller the heat of combustion and the efficiency, the larger the mass. The model is simple: the efficiency of the boiler is assumed to stay constant.

   Formula

   $E_\text{useful} = \eta H m$, so $m = \dfrac{E_\text{useful}}{\eta H}$, and $\Delta m = m_\text{wood} - m_\text{pellet}$

   Conversions

   Not needed: the energy is in megajoules and the heats of combustion in megajoules per kilogram. The percentages as ratios are $70\,\% = 0.70$, $85\,\% = 0.85$, $90\,\% = 0.90$ and $75\,\% = 0.75$.

   Substitution

   $m_\text{wood}$$= \dfrac{36\,000\,\text{MJ}}{0.70 \cdot 15\,\tfrac{\text{MJ}}{\text{kg}}}$

   $= 3429\,\text{kg}$

   $m_\text{pellet}$$= \dfrac{36\,000\,\text{MJ}}{0.85 \cdot 17\,\tfrac{\text{MJ}}{\text{kg}}}$

   $= 2491\,\text{kg}$

   $m_\text{oil}$$= \dfrac{36\,000\,\text{MJ}}{0.90 \cdot 43\,\tfrac{\text{MJ}}{\text{kg}}}$

   $= 930\,\text{kg}$

   The difference is

   $\Delta m$$= 3429\,\text{kg} - 2491\,\text{kg}$

   $= 938\,\text{kg}$

   At efficiency $0.75$ the pellets used are

   $m_\text{pellet}'$$= \dfrac{36\,000\,\text{MJ}}{0.75 \cdot 17\,\tfrac{\text{MJ}}{\text{kg}}}$

   $= 2824\,\text{kg}$

   Answer

   in the answer box, the result of part b), $940\,\text{kg}$.
   a) The wood used is $3400\,\text{kg}$, the pellets $2500\,\text{kg}$ and the oil $930\,\text{kg}$.
   b) About $940\,\text{kg}$ more wood than pellets is used, two significant figures like the data $36\,000\,\text{MJ}$.
   c) The consumption of pellets rises at efficiency $0.75$ to $2800\,\text{kg}$, that is $330\,\text{kg}$ more: when the efficiency drops, part of the energy released goes to losses and more fuel is needed for the same useful energy.
   d) The least oil is used, because its heat of combustion is more than twice that of wood and pellets ($43 : 15 = 2.9$), while the efficiencies are of the same order of magnitude ($0.90$ and $0.70$); the heat of combustion determines the order of magnitude of the mass.

   Reasonableness

   An estimate without losses: wood $36\,000 : 15 = 2400\,\text{kg}$ and pellets $36\,000 : 17 = 2120\,\text{kg}$. The efficiencies increase these to $2400 : 0.70 = 3430\,\text{kg}$ and $2120 : 0.85 = 2490\,\text{kg}$, which agrees with the result. The mass for the pellet boiler grows by a factor of $2824\,\text{kg} : 2491\,\text{kg} = 1.13$, and $0.85 : 0.75 = 1.13$, as it should.