{"id":72,"date":"2020-05-13T10:45:09","date_gmt":"2020-05-13T14:45:09","guid":{"rendered":"https:\/\/opentextbc.ca\/mcluhantest\/?post_type=chapter&p=72"},"modified":"2022-11-21T10:52:09","modified_gmt":"2022-11-21T15:52:09","slug":"heat-energy","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/mcluhantest\/chapter\/heat-energy\/","title":{"raw":"Heat Energy Measurements","rendered":"Heat Energy Measurements"},"content":{"raw":"
This chapter was removed at the request of Mark (and supported by Tim) after Josie received an error report that detailed multiple inaccuracies with the content. Mark believed those corrections would make the content more complicated than he wanted for a math book, and stated that removing the chapter would not disrupt the flow of the content. This content may be useful in a future science textbook once it is updated.<\/span><\/div>\n\"Image\n\nTake a look at the picture of the fire to the left. If you were to describe that fire and you had the three words heat, temperature and energy to choose from, which would you say best describes the fire?\n\nYou might say that all three words could describe the fire and you would be right. As all three words can be used to describe a similar situation this could be confusing. As this chapter deals with heat energy you might have guessed that we are going look at this from the heat energy point of view. We\u2019ll get to temperature in the next section but for now know that temperature is a measure of the intensity of the heat. We could describe the fire as very hot which would indicate a high intensity of heat.\n\nAt this point we should get a definition of energy. I\u2019m going to refer to WIKIPEDIA here for the definition:\n\nEnergy:<\/strong> In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object.\n\nThanks Wikipedia!\n\nThere are many forms of energy. We are just going to focus on heat energy in this section.\n

\"\"<\/p>\n

If you would like to know more about the different forms of energy check out the following link: Types of Energy (Solar Schools)<\/a><\/p>\n\n

Measuring Heat Energy<\/h1>\nIn the last two sections we started off by going through the metric units and then the imperial units. Well there are only two units for each that we are going to deal with so we\u2019ll introduce them all at once and then work through examples within metric and within imperial and then work between them. The four units we are going to deal with are as follows:\n\n\n\n\n\n
Metric<\/th>\nImperial<\/th>\n<\/tr>\n
kilowatts (kW)<\/td>\nBritish thermal unit (btu)<\/td>\n<\/tr>\n
calories (cal)<\/td>\njoules (J)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nLet's take a look at the relationship between each of the units.\n\n\n\n\n\n\n\n
Unit<\/th>\nEquivalent<\/th>\n<\/tr>\n
kilowatt<\/td>\n3412 btu's\u00a0 \u00a0 860,421 cal<\/td>\n<\/tr>\n
British thermal unit<\/td>\n0.293 kW\u00a0 \u00a01055 J<\/td>\n<\/tr>\n
calorie<\/td>\n4.186 J<\/td>\n<\/tr>\n
joules<\/td>\n0.239 cal<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nNote that there are a couple of equivalent numbers missing such as the number of kilowatts in a joule. These are not included as they are either too small or way large to bother with. The numbers above are generally the ones we need to know.\n\nSomething else to note is that often instead of using calories and joules we use kilocalories and kilojoules. A kilocalorie (kcal) is 1000 calories and it takes 1 kilocalorie to raise the temperature of 1 kilogram of water 1\u00b0C. As we learned before there are 1000 grams in one kilogram. Following the logic it would then take 1 calorie to raise 1 gram of water 1\u00b0C.\n

[latex] \\begin{array}{rcl} \\text{1 kilocalorie (1000 calories)} & \\longrightarrow & \\text{raises 1 kilogram (1000 grams) of water 1\u00b0C} \\\\ \\text{1 calorie}& \\longrightarrow & \\text{raises 1 gram of water 1\u00b0C} \\end{array}[\/latex]<\/p>\nThis same logic follows for the joule. One kilojoule equals 1000 joules. If we relate it to increasing the temperature of water what we would see is that is takes 1 joule to raise 1 gram of water 0.24\u00b0C. Following the same logic as the calorie and kilocalorie it would take 1 kilojoule to raise 1 kilogram of water 0.24\u00b0C.\n

[latex] \\begin{array}{rcl} \\text{1 kilojoule (1000 joules)} & \\longrightarrow & \\text{raises 1 kilogram (1000 grams) of water 0.24\u00b0C} \\\\ \\text{1 joule}& \\longrightarrow & \\text{raises 1 gram of water 0.24\u00b0C} \\end{array}[\/latex]<\/p>\nWe can take this a bit further and talk about the heat energy in a British thermal unit. The heat energy in one BTU is enough to raise one pound of water 1 degree Fahrenheit. It\u2019s generally considered to be the amount of heat energy in a match.\n

\n\n\"Note.\"Remember that you can derive any other number you need from the conversion numbers given. These are the only number that you have to memorize.\n\n \n\n<\/div>\nLet\u2019s go through a few example questions to see where we are at.\n
\n

Example<\/p>\n\n<\/header>\n

\n\nHow many joules (J) are there in 14 British thermal units (BTU)?\n\nStep 1: <\/strong>Find the number that translates between joules and British thermal units. Keep in mind that we are going from BTU\u2019s to joules.\n\nIn this case we know that 1 BTU = 1055 joules\n\nStep 2: <\/strong>As usual build a ratio.\n

[latex]\\Large \\dfrac{\\text {1 BTU}}{\\text{14 BTU's}} = \\dfrac{\\text{1055 joules}}{\\text{X joules}}[\/latex]<\/p>\nStep 3:<\/strong> Cross multiply.\n

[latex]\\Large\\begin{array}{c} \\dfrac{\\text {1 BTU}}{\\text{14 BTU's}} = \\dfrac{\\text{1055 joules}}{\\text{X joules}} \\\\ 1 \\times \\text{X} = 14 \\times 1055 \\\\ \\text{X} = 14,470 \\\\ \\text{Answer} = 14,470 \\text{ joules}\\end{array}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n

\n

Example<\/p>\n\n<\/header>\n

\n\nHow many kilowatts (kW) are there in 1,495, 276 calories (cal)?\n\nStep 1: <\/strong>Find the number that translates between kilowatts and calories.\n\nIn this case we know that 1 kilowatt is equal to 860,421 calories. Note that we are going from calories to kilowatts and we don\u2019t have\u00a0 the number of kW in one calorie. What we do have is the number to go from kilowatts to calories. We know that 1 kilowatt = 860,421 calories.\n\nStep 2: <\/strong>Build a ratio using the number we do have.\n

[latex]\\Large \\dfrac{\\text {1 kW}}{\\text{X kW}} = \\dfrac{\\text{860,421 cal}}{\\text{1,495,276 cal}}[\/latex]<\/p>\nWhat you'll note here is that when you get to the stage where you cross multiply the equation doesn't end up as nice and easy to work with as we've had so far. We don't end up with the \"1 x X.\" What we have to do here is manipulate the equation to solve for X. This is a little out of the scope of what we have gone through so far but you'll find a full explanation in the next chapter. Having said all that let's go through the motions to solve the equation.\n\nStep 3:<\/strong> Cross multiply\n

[latex]\\Large\\begin{array}{c}\\dfrac{\\text {1 kW}}{\\text{X kW}} = \\dfrac{\\text{860,421 cal}}{\\text{1,495,276 cal}} \\\\ 1 \\times 1,495,276 = \\text{X} \\times 860,421 \\\\ \\text{X} = \\dfrac{1,495,276}{860,421}= 1.738 \\\\ \\text{Answer} = 1.738 \\text{ kW}\\end{array} [\/latex]<\/p>\n\n<\/div>\n<\/div>\nNow try a question for yourself\n

Practice Question<\/h1>\nTry a practice question yourself and check the video answer to see how you did. Make sure to follow the steps outlined above and think about whether your answer should be bigger or smaller.\n
\n

Question 1<\/p>\n\n<\/header>\n

\n\nAmir and Parviz are installing a boiler. The boiler is rated at 110 kW. Their gas fitting ticket allows them to install and fire appliances up to and including 400,000 BTU\u2019s per hour. Are they allowed to fire this appliance?\n\nhttps:\/\/media.bccampus.ca\/id\/0_j4gtnoqx?width=608&height=402&playerId=23449753\n\n<\/div>\n<\/div>","rendered":"
This chapter was removed at the request of Mark (and supported by Tim) after Josie received an error report that detailed multiple inaccuracies with the content. Mark believed those corrections would make the content more complicated than he wanted for a math book, and stated that removing the chapter would not disrupt the flow of the content. This content may be useful in a future science textbook once it is updated.<\/span><\/div>\n

\"Image<\/p>\n

Take a look at the picture of the fire to the left. If you were to describe that fire and you had the three words heat, temperature and energy to choose from, which would you say best describes the fire?<\/p>\n

You might say that all three words could describe the fire and you would be right. As all three words can be used to describe a similar situation this could be confusing. As this chapter deals with heat energy you might have guessed that we are going look at this from the heat energy point of view. We\u2019ll get to temperature in the next section but for now know that temperature is a measure of the intensity of the heat. We could describe the fire as very hot which would indicate a high intensity of heat.<\/p>\n

At this point we should get a definition of energy. I\u2019m going to refer to WIKIPEDIA here for the definition:<\/p>\n

Energy:<\/strong> In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object.<\/p>\n

Thanks Wikipedia!<\/p>\n

There are many forms of energy. We are just going to focus on heat energy in this section.<\/p>\n

\"\"<\/p>\n

If you would like to know more about the different forms of energy check out the following link: Types of Energy (Solar Schools)<\/a><\/p>\n

Measuring Heat Energy<\/h1>\n

In the last two sections we started off by going through the metric units and then the imperial units. Well there are only two units for each that we are going to deal with so we\u2019ll introduce them all at once and then work through examples within metric and within imperial and then work between them. The four units we are going to deal with are as follows:<\/p>\n\n\n\n\n\n
Metric<\/th>\nImperial<\/th>\n<\/tr>\n
kilowatts (kW)<\/td>\nBritish thermal unit (btu)<\/td>\n<\/tr>\n
calories (cal)<\/td>\njoules (J)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let’s take a look at the relationship between each of the units.<\/p>\n\n\n\n\n\n\n\n
Unit<\/th>\nEquivalent<\/th>\n<\/tr>\n
kilowatt<\/td>\n3412 btu’s\u00a0 \u00a0 860,421 cal<\/td>\n<\/tr>\n
British thermal unit<\/td>\n0.293 kW\u00a0 \u00a01055 J<\/td>\n<\/tr>\n
calorie<\/td>\n4.186 J<\/td>\n<\/tr>\n
joules<\/td>\n0.239 cal<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Note that there are a couple of equivalent numbers missing such as the number of kilowatts in a joule. These are not included as they are either too small or way large to bother with. The numbers above are generally the ones we need to know.<\/p>\n

Something else to note is that often instead of using calories and joules we use kilocalories and kilojoules. A kilocalorie (kcal) is 1000 calories and it takes 1 kilocalorie to raise the temperature of 1 kilogram of water 1\u00b0C. As we learned before there are 1000 grams in one kilogram. Following the logic it would then take 1 calorie to raise 1 gram of water 1\u00b0C.<\/p>\n

[latex]\\begin{array}{rcl} \\text{1 kilocalorie (1000 calories)} & \\longrightarrow & \\text{raises 1 kilogram (1000 grams) of water 1\u00b0C} \\\\ \\text{1 calorie}& \\longrightarrow & \\text{raises 1 gram of water 1\u00b0C} \\end{array}[\/latex]<\/p>\n

This same logic follows for the joule. One kilojoule equals 1000 joules. If we relate it to increasing the temperature of water what we would see is that is takes 1 joule to raise 1 gram of water 0.24\u00b0C. Following the same logic as the calorie and kilocalorie it would take 1 kilojoule to raise 1 kilogram of water 0.24\u00b0C.<\/p>\n

[latex]\\begin{array}{rcl} \\text{1 kilojoule (1000 joules)} & \\longrightarrow & \\text{raises 1 kilogram (1000 grams) of water 0.24\u00b0C} \\\\ \\text{1 joule}& \\longrightarrow & \\text{raises 1 gram of water 0.24\u00b0C} \\end{array}[\/latex]<\/p>\n

We can take this a bit further and talk about the heat energy in a British thermal unit. The heat energy in one BTU is enough to raise one pound of water 1 degree Fahrenheit. It\u2019s generally considered to be the amount of heat energy in a match.<\/p>\n

\n

\"Note.\"Remember that you can derive any other number you need from the conversion numbers given. These are the only number that you have to memorize.<\/p>\n

 <\/p>\n<\/div>\n

Let\u2019s go through a few example questions to see where we are at.<\/p>\n

\n
\n

Example<\/p>\n<\/header>\n

\n

How many joules (J) are there in 14 British thermal units (BTU)?<\/p>\n

Step 1: <\/strong>Find the number that translates between joules and British thermal units. Keep in mind that we are going from BTU\u2019s to joules.<\/p>\n

In this case we know that 1 BTU = 1055 joules<\/p>\n

Step 2: <\/strong>As usual build a ratio.<\/p>\n

[latex]\\Large \\dfrac{\\text {1 BTU}}{\\text{14 BTU's}} = \\dfrac{\\text{1055 joules}}{\\text{X joules}}[\/latex]<\/p>\n

Step 3:<\/strong> Cross multiply.<\/p>\n

[latex]\\Large\\begin{array}{c} \\dfrac{\\text {1 BTU}}{\\text{14 BTU's}} = \\dfrac{\\text{1055 joules}}{\\text{X joules}} \\\\ 1 \\times \\text{X} = 14 \\times 1055 \\\\ \\text{X} = 14,470 \\\\ \\text{Answer} = 14,470 \\text{ joules}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

\n
\n

Example<\/p>\n<\/header>\n

\n

How many kilowatts (kW) are there in 1,495, 276 calories (cal)?<\/p>\n

Step 1: <\/strong>Find the number that translates between kilowatts and calories.<\/p>\n

In this case we know that 1 kilowatt is equal to 860,421 calories. Note that we are going from calories to kilowatts and we don\u2019t have\u00a0 the number of kW in one calorie. What we do have is the number to go from kilowatts to calories. We know that 1 kilowatt = 860,421 calories.<\/p>\n

Step 2: <\/strong>Build a ratio using the number we do have.<\/p>\n

[latex]\\Large \\dfrac{\\text {1 kW}}{\\text{X kW}} = \\dfrac{\\text{860,421 cal}}{\\text{1,495,276 cal}}[\/latex]<\/p>\n

What you’ll note here is that when you get to the stage where you cross multiply the equation doesn’t end up as nice and easy to work with as we’ve had so far. We don’t end up with the “1 x X.” What we have to do here is manipulate the equation to solve for X. This is a little out of the scope of what we have gone through so far but you’ll find a full explanation in the next chapter. Having said all that let’s go through the motions to solve the equation.<\/p>\n

Step 3:<\/strong> Cross multiply<\/p>\n

[latex]\\Large\\begin{array}{c}\\dfrac{\\text {1 kW}}{\\text{X kW}} = \\dfrac{\\text{860,421 cal}}{\\text{1,495,276 cal}} \\\\ 1 \\times 1,495,276 = \\text{X} \\times 860,421 \\\\ \\text{X} = \\dfrac{1,495,276}{860,421}= 1.738 \\\\ \\text{Answer} = 1.738 \\text{ kW}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

Now try a question for yourself<\/p>\n

Practice Question<\/h1>\n

Try a practice question yourself and check the video answer to see how you did. Make sure to follow the steps outlined above and think about whether your answer should be bigger or smaller.<\/p>\n

\n
\n

Question 1<\/p>\n<\/header>\n

\n

Amir and Parviz are installing a boiler. The boiler is rated at 110 kW. Their gas fitting ticket allows them to install and fire appliances up to and including 400,000 BTU\u2019s per hour. Are they allowed to fire this appliance?<\/p>\n