Temperature

5608 /www/nrich/html/content/id/5608/ 1 2011-03-09T12:41:23 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/6951">Warm-up</a></li> <li><a href="http://nrich.maths.org/996">Try this next</a></li> <li><a href="http://nrich.maths.org/2032">Think higher</a></li> <li><a href="http://lamar.colostate.edu/~hillger/temps.htm">Read: mathematics</a></li> <li><a href="http://cryo.gsfc.nasa.gov/introduction/temp_scales.html">Read: science</a></li> <li><a href="http://plus.maths.org/content/met-office-another-roasting">Explore further</a></li> </ul> <div> </div> Temperature is often measured in degrees Celsius,$^\circ C$, or degrees Fahrenheit,$^\circ F$. <mdo:image alt="" src="thermometer%20new.PNG" style="width: 122px; height: 528px; float: right;"></mdo:image> The freezing point of water is $0^\circ$C and $32^\circ$F. The boiling point of water is $100^\circ$C and $212^\circ$F. Is there a temperature at which Celsius and Fahrenheit readings are the same? Can you describe a way of converting Fahrenheit readings into Celsius? Can you describe a way of converting Celsius readings into Fahrenheit? <div> </div> <div> </div> An extension challenge: Scientists often use the Kelvin scale of temperature, where the freezing point of water is $273.15^\circ K$ and the boiling point of water is $373.15^\circ K$. Is there a temperature at which Kelvin and Fahrenheit readings are the same? Is there a temperature at which Kelvin and Celsius readings are the same? Can you describe ways of converting Kelvin readings into Fahrenheit and Celsius readings? <a href="http://nrich.maths.org/7211">Click here for a poster of this problem</a>.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Students from Ms White's Yr 7 maths class in Lyneham High School, Australia, noticed that: Every time you go down 5 degrees in the Celsius scale, you go down 9 degrees in the Fahrenheit scale. 0ºC = 32ºF -5ºC = 32ºF - 9 = 23ºF -10ºC = 23ºF - 9 = 14ºF -15ºC = 14ºF - 9 = 5ºF -20ºC = 5ºF - 9 = -4ºF -25ºC = -4ºF - 9 = -13ºF -30ºC = -13ºF - 9 = -22ºF -35ºC = -22ºF - 9 = -31 ºF -40ºC = -31ºF - 9 = -40ºF -40ºC = -40ºF They described a way of converting Celsius readings into Fahrenheit: We know 0 degrees Celsius = 32 degrees Fahrenheit. And an interval of 1 degree Celsius = 1.8 degrees Fahrenheit. So, for example, if you want 5 degrees Celsius, you have to times 1.8 by 5 and then add 32. C x 1.8 + 32 = F And then rearranged this formula to convert Fahrenheit into Cesius: You can reverse the equation above and say (F - 32) ÷ 1.8 = C This is how they found a temperature at which the Fahrenheit reading is 20 degrees higher than the Celsius reading: C x 1.8 + 32 = F and F = C + 20 C x 1.8 + 32 = C + 20 32 + 0.8 C = 20 0.8 C = -12 C = -15 F = 5 Nina, Kristjan and Jure from the Elementary School in Loka Crnomelj in Slovenia sent us <a href="/content/id/5608/SloveniaTemperature.pdf" class="editorial">this</a> most comprehensive solution, explaining clearly all stages of their work. Many students knew the formula for converting Celsius into Fahrenheit and used this to find the temperature at which the Celsius and Fahrenheit readings are the same. <a href="/content/id/5608/MichaelTemperature.doc" class="editorial">Here</a> are the three strategies that Michael from Bilton School used to find the temperature. This is how Samuel from Long Buckby Junior School reasoned: There is a temperature at which Celsius and Farenheit are the same. It is $-40$ degrees, because $9/5$ of $-40$ is $-72$ and $-72 + 32 = -40$. I decided to look at negative numbers because starting with a positive number and multiplying it by $9/5$ is going to increase it and so is adding 32 so you're always going to end up with a number greater than the number you started with. However, if you start with a negative number, multiplying it by $9/5$ decreases it, and adding $32$ increases it, so I realised that with the correct number, Celsius and Fahrenheit might be the same. I decided to go down in tens: $9/5$ of $-10 = -18$ and $-18 + 32 = 14$, so that doesn't work; $9/5$ of $-20 = -36$ and $-36 + 32 = -4$, so that doesn't work; $9/5$ of $-30 = -54$ and $-54 + 32 = -22$, so that doesn't work. But $9/5$ of $-40 = -72$ and $-72 + 32 = -40$ so it works. The reason it works is because multiplying by $9/5$ is equivalent to adding $4/5$ of it, and for $-40$ adding $32$ is equivalent to subtracting $4/5$ of it (because $32$ is $4/5$ of $40$). Because of this, Farenheit and Celsius are equivalent ONLY at $-40$ degrees. The Four Mathemateers from Brocks Hill Primary School also used a trial and error approach, as displayed here: First we started going down in tens of Celsius from $0$, and we found out a pattern: the difference between F and C was getting closer by eights every time. When we got to $-30C$ the difference was only $8$. So $-30$C is equal to $-22$F. Then we tried $-40$C and found out that $-40$C was the same as $-40$F. So the answer is $-40$. Yesuhei used a similar strategy: <div>First I tried different solutions for Celsius like $-50$ and that gave me $-58$ Fahrenheit .</div> <div>Then I tried $-45$C because whenever you go down (negative increasing) the F and C's distance increases. That gave me $-48$ Fahrenheit which was very close, so I tried $-40$ Celsius and that gave me $-40$ Fahrenheit.</div> Others who found the correct answer by this method are Emma and Chloe from The Mount School and Michael from Bilton School. Beatrice fron Raffles Girls' School and Michael used a graphical approach. Michael's answer is shown here: I plotted the lines of the simultaneous equations against each other and found where they crossed. In the graphs $y =$ F and $x = $C. <mdo:image width="610" height="450" src="graph.gif" alt="graph"></mdo:image> The quickest way to solve this problem is with an algebraic approach, and both of the people who used graphs used this approach as well. The other people that obtained the correct answer by this method include Vyas from Wilson's School, Patrick from Woodbridge School, Sugam and Fiona from The Mount School, Chris from CCSN, Samantha from The Steele School, Gemma, Griselda and Charlie from Colyton Grammar School, Jasvir, Matt and Christian from Kingshill, Ed from Tunbridge Wells Grammar School for Boys, Stephen and Joe from Singapore International School, Kieran from Alcester Grammar School and Pradeesha. Here is the solution from Vyas: F = 9C/5 + 32 When C = F C = 9C/5 + 32 5C = 9C + 160 4C = -160 C = -40 Here is how Patrick solved the problem: Since F = (9C/5) + 32 then F = 9F/5 + 32 therefore F - 32 = 9F/5 5F - 160 = 9F 4F = -160 F = -40 To check: -40 = [(9 x -40) / 5] + 32 -40 = (-360 / 5) + 32 -40 = -72 + 32 -40 = -40 QED Here is Sugam's working: Let x be the temperature where Fahranheit and Celsius are equal. $x=\frac{9}{5} x+32$ $5x = 9x + 160 $ $-4x = 160 $ $x = -40 $ Therefore $-40$ Celsius $= -40$ Farenheit And here is Michael's solution: To solve it algebraically I can create two simultaneous equations: $F = C$ $F = 1.8C + 32 $ Therefore $C = 1.8C + 32 $ $C = -32 / 0.8 = -40 $ And here is Kieran's solution: Using the equation, $F=\frac{9}{5} C+32$, we can remove the fraction by multiplying both sides by five. Doing so produces $5F = 9C + 160$, and thus, using the sought after equation of $F = C$, we may further deduce that, since $5F$ and $5C$ are one and the same, subtracting the two equal amounts from either side leaves $0 = 4C + 160$ or $4C = -160$ or $C = -40$ Consequently, $-40$ Celcius is the same as $-40$ Fahrenheit. Oliver remembered to check that his solution worked: We can substitute $-40$ as $C$ in $F = 9/5C + 32$ to check our answer As $-40 = -72 + 32$, our answer is correct Beatrice combined an algebraic and graphical approach: <div>We know that $F = (9/5)C + 32$ [equation 1]</div> This is a linear equation, because it follow the structure $y = mx + c$ Now let us make C the subject: $F = (9/5)C + 32 $ $F - 32 = (9/5)C$ $C = (5/9)(F - 32)$ [equation 2] Now plot equations 1and 2 on Graphmatica. They will intersect at the point $(-40,-40)$. So we know that $-40F = -40C$. <mdo:image width="633" height="480" alt="graph" src="grapg2.gif"></mdo:image> Patrick from Otterbourne was interested in finding temperatures at which readings are the same, for Celsius and Fahrenheit, Kelvin and Celsius, and Fahrenheit and Kelvin. <a href="/content/id/5608/Patrick-Temperature.doc" class="editorial">Here</a> is his solution. Alexander from Wilson's School was interested in finding how to convert Celsius and Fahrenheit readings into Kelvin: A good way of converting Celsius into Kelvin is to just add 273.15 to Celsius and you will get Kelvin. This is how I worked it out: Celsius temperatures: Freezing: 0C Boiling: 100C. Total gap: 100. Kelvin temperatures: Freezing: 273.15K Boiling: 373.15K Total gap: 100. Therefore, because the gap is the same, you just have to add 273.15 to the Celsius temperature to get Kelvin, and vice-versa. To convert Fahrenheit into Kelvin you need to do the following equation: Kelvin = [(Fahrenheit - 32) x 5/9] + 273.15 The way I found this was that I did the formula from Fahrenheit into Celsius and added 273.15 to the end. Vatsal, also from Wilson's School, worked on finding temperatures at which the Celsius and Fahrenheit readings differed by 20: The formula to change Farenheit (F) into Celcius (C) is: F = 9C/5 + 32 To find at what temperature Farenheit and Celcius are the same, you would do this: F = 9F/5 + 32 To find at what temperature Farenheit is 20 higher than Celcius, you would do this: F = 9(F - 20)/5 + 32 F = (9F - 180)/5 + 32 F - 32 = (9F - 180)/5 5F - 160 = 9F - 180 -4F= -20 F = 5 This tells us that 5 degrees Farenheit is the same as -15 degrees Celcius. To find at what temperature Celcius is 20 higher than Farenheit, you would do this: C - 20 = 9C/5 + 32 Niharika from Leicester High School for Girls sent us <a href="/content/id/5608/Niharika-Temperature.pdf" class="editorial">this</a> comprehensive solution to all the different questions in the problem. Well done to all of you who contributed solutions to this problem. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=5608&part="> This problem</a> offers an opportunity to combine skills from mathematics and science. It can be solved numerically, algebraically or graphically, so can offer a useful opportunity for discussing the merits of different methods.</div> <h3>Possible approach</h3> <div>Introduce the boiling and freezing point of water in Celsius and Fahrenheit.</div> <div>"What other information can you deduce from these temperature facts?"</div> <div>Give the class some time to discuss in pairs, then bring the class together to collect ideas on the board. Possible responses might be:</div> <div>"$50^\circ C = 122^\circ F$ because it's halfway between."</div> <div>"$200^\circ C = 392^\circ F$ because it's another $180^\circ F$."</div> <div>"A temperature increase of $100^\circ C$ is the same as a temperature increase of $180^\circ F$."</div> <div> </div> <div>"Can you use everyone's ideas to deduce any more information about the temperature scales?"</div> <div>Again, give the class some time to discuss in pairs, and then collect ideas once more.</div> <div> </div> <div>"Is there a temperature where the reading in Celsius is the same as the reading in Fahrenheit?"</div> <div>Give the class plenty of time to approach this problem. Most students are likely to use a numerical approach. If some students use algebraic or graphical methods, ask them to share their approaches with the rest of the class.</div> <div> </div> <div>If nobody uses algebra or graphs, ask the class to consider first how a graph might help:</div> <div>"Can you represent the original information graphically in a way that could have helped you to solve the problem?"</div> <div> </div> <div>The graphical method can then lead on to a discussion of the algebraic representation of the straight line graph and hence algebraic methods of solution.</div> <div> </div> <div>Take time to discuss the merits of the different methods and then challenge students to show how to use each solution method to solve problems such as:</div> <div>"Is there a temperature at which the Fahrenheit reading is 20 degrees higher than the Celsius reading?"</div> <div>"Is there a temperature at which the Celsius reading is 20 degrees higher than the Fahrenheit reading?" </div> <h3>Key questions</h3> <div>Does every method give the same answer?</div> <div>What are the merits of the different methods?</div> <h3>Possible extension</h3> See the extension challenge introducing the Kelvin scale of temperature in the problem. <h3>Possible support</h3> Spend lots of time discussing how to deduce information from the initial temperature facts given. Perhaps it would help students to suggest new values if the information is presented in a table. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">You could use a graphical method... You could try an algebraic approach... </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">-15C is 5 F -65C is -85F The answer to this problem is yes, there IS a temperature where Celsius and Fahrenheit are equal, and it is $-40$ degrees. All the solutions to this problem took one of three forms: trial-and-improvement, graphical or algebraic: This is how Samuel from Long Buckby Junior School reasoned: There is a temperature at which Celsius and Farenheit are the same. It is $-40$ degrees, because $9/5$ of $-40$ is $-72$ and $-72 + 32 = -40$. I decided to look at negative numbers because starting with a positive number and multiplying it by $9/5$ is going to increase it and so is adding 32 so you're always going to end up with a number greater than the number you started with. However, if you start with a negative number, multiplying it by $9/5$ decreases it, and adding $32$ increases it, so I realised that with the correct number, Celsius and Fahrenheit might be the same. I decided to go down in tens: $9/5$ of $-10 = -18$ and $-18 + 32 = 14$, so that doesn't work; $9/5$ of $-20 = -36$ and $-36 + 32 = -4$, so that doesn't work; $9/5$ of $-30 = -54$ and $-54 + 32 = -22$, so that doesn't work. But $9/5$ of $-40 = -72$ and $-72 + 32 = -40$ so it works. The reason it works is because multiplying by $9/5$ is equivalent to adding $4/5$ of it, and for $-40$ adding $32$ is equivalent to subtracting $4/5$ of it (because $32$ is $4/5$ of $40$). Because of this, Farenheit and Celsius are equivalent ONLY at $-40$ degrees. The Four Mathemateers from Brocks Hill Primary School also used a trial and error approach, as displayed here: First we started going down in tens of Celsius from $0$, and we found out a pattern: the difference between F and C was getting closer by eights every time. When we got to $-30C$ the difference was only $8$. So $-30$C is equal to $-22$F. Then we tried $-40$C and found out that $-40$C was the same as $-40$F. So the answer is $-40$. Yesuhei used a similar strategy: <div>First I tried different solutions for Celsius like $-50$ and that gave me $-58$ Fahrenheit .</div> <div>Then I tried $-45$C because whenever you go down (negative increasing) the F and C's distance increases. That gave me $-48$ Fahrenheit which was very close, so I tried $-40$ Celsius and that gave me $-40$ Fahrenheit.</div> Others who found the correct answer by this method are Emma and Chloe from The Mount School and Michael from Bilton School. Beatrice fron Raffles Girls' School and Michael used a graphical approach. Michael's answer is shown here: I plotted the lines of the simultaneous equations against each other and found where they crossed. In the graphs $y =$ F and $x = $C. <mdo:image alt="graph" height="450" src="graph.gif" width="610"></mdo:image> The quickest way to solve this problem is with an algebraic approach, and both of the people who used graphs used this approach as well. The other people that obtained the correct answer by this method include Sugam and Fiona from The Mount School, Chris from CCSN, Samantha from The Steele School, Gemma, Griselda and Charlie from Colyton Grammar School, Jasvir, Matt and Christian from Kingshill, Ed from Tunbridge Wells Grammar School for Boys, Stephen and Joe from Singapore International School, Kieran from Alcester Grammar School and Pradeesha: Here is Sugam's working: Let x be the temperature where Fahranheit and Celsius are equal. $x=\frac{9}{5} x+32$ $5x = 9x + 160 $ $-4x = 160 $ $x = -40 $ Therefore $-40$ Celsius $= -40$ Farenheit And here is Michael's solution: To solve it algebraically I can create two simultaneous equations: $F = C$ $F = 1.8C + 32 $ Therefore $C = 1.8C + 32 $ $C = -32 / 0.8 = -40 $ And here is Kieran's solution: Using the equation, $F=\frac{9}{5} C+32$, we can remove the fraction by multiplying both sides by five. Doing so produces $5F = 9C + 160$, and thus, using the sought after equation of $F = C$, we may further deduce that, since $5F and 5C$ are one and the same, subtracting the two equal amounts from either side leaves $0 = 4C + 160$ or $4C = -160$ or $C = -40$ Consequently, $-40$ Celcius is the same as $-40$ Fahrenheit. Oliver remembered to check that his solution worked: We can substitute $-40$ as $C$ in $F = 9/5C + 32$ to check our answer As $-40 = -72 + 32$, our answer is correct Beatrice combined an algebraic and graphical approach: <div>We know that $F = (9/5)C + 32$ [equation 1]</div> This is a linear equation, because it follow the structure $y = mx + c$ Now let us make C the subject: $F = (9/5)C + 32 $ $F - 32 = (9/5)C$ $C = (5/9)(F - 32)$ [equation 2] Now plot equations 1and 2 on Graphmatica. They will intersect at the point $(-40,-40)$. So we know that $-40F = -40C$. <mdo:image alt="graph" height="480" src="grapg2.gif" width="633"></mdo:image> Well done to all of you who solved this problem correctly. Short solution: 1. Is there a temperature at which Celsius and Fahrenheit readings are the same? Yes, -40°C = -40°F 2. Can you describe a way of converting Fahrenheit readings into Celsius? $C = (F-32):1.8$ where C is a temperature in Celsius and F is the temperature in Fahrenheit. 3. Can you describe a way of converting Celsius readings into Fahrenheit? $F = 32+1.8C$ 4. Is there a temperature at which the Fahrenheit reading is <span class="mrow" id="MathJax-Span-49" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; display: inline; position: static; border-top-width: 0px; border-right-width: 0px; border-bottom-width: 0px; border-left-width: 0px; border-style: initial; border-color: initial; vertical-align: 0px; line-height: normal; text-decoration: none; font-family: MathJax_Main, MathJax_Size1, MathJax_AMS; "> <span class="mn" id="MathJax-Span-50" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; display: inline; position: static; border-top-width: 0px; border-right-width: 0px; border-bottom-width: 0px; border-left-width: 0px; border-style: initial; border-color: initial; vertical-align: 0px; line-height: normal; text-decoration: none; font-family: MathJax_Main; "> 20 degrees higher than the Celsius reading? Yes, -15°C = 5°F 5. Is there a temperature at which the Celsius reading is <span class="mrow" id="MathJax-Span-52" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; display: inline; position: static; border-top-width: 0px; border-right-width: 0px; border-bottom-width: 0px; border-left-width: 0px; border-style: initial; border-color: initial; vertical-align: 0px; line-height: normal; text-decoration: none; font-family: MathJax_Main, MathJax_Size1, MathJax_AMS; "> <span class="mn" id="MathJax-Span-53" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; display: inline; position: static; border-top-width: 0px; border-right-width: 0px; border-bottom-width: 0px; border-left-width: 0px; border-style: initial; border-color: initial; vertical-align: 0px; line-height: normal; text-decoration: none; font-family: MathJax_Main; "> 20 degrees higher than the Fahrenheit reading? Yes, -65°C = -85°F 6. Is there a temperature at which Kelvin and Fahrenheit readings are the same? Yes, 574.6°K = 574.6°F 7. Is there a temperature at which Kelvin and Celsius readings are the same? No 8. Can you describe ways of converting Kelvin readings into Fahrenheit and Celsius readings? F = 32 + 1.8(K-273.15) and C = K - 273.15 where K is the temperature in Kelvin.</mdoxml> 2 4 0 0 1 0 0 Temperature Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same? Algebra Manipulating algebraic expressions/formulae Algebra Solving equations graphically Algebra Linear equations Fractions, Decimals, Percentages, Ratio and Proportion Calculating with ratio & proportion Information and Communications Technology smartphone Applications STEM - General Applications Maths Supporting SET Applications STEM - physical world Admin Discussion Admin Individual Secondary Mapping Document Equations and formulae LS Secondary Mapping Document DisplayCabinet