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Heat Practice Problems With Detailed Answers

Struggling with heat problems in your physics class?

Our article is here to help! Packed with solved examples specifically designed for high school students, this guide will make mastering heat problems easy.

Get ready to boost your grades and deepen your understanding with these easy-to-follow solutions!

When heat energy $Q$ causes a change in temperature $\Delta T=T_f-T_i$ in a sample with specific heat capacity $c$ and mass $m$, then we can relate all these physical quantities as following formula \[Q=mc\Delta T=mc(T_f-T_i)\] where $T_f$ and $T_i$ are the initial and final temperatures. 

Heat Practice Problems

Problem (1):  5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 0.092 cal/g. °C) 

Solution : The energy required to change the temperature of a substance of mass $m$ from initial temperature $T_i$ to final temperature $T_f$ is obtained by the formula $Q=mc(T_f-T_i)$, where $c$ is the specific heat of the substance. Thus, we have \begin{align*} Q&=mc\Delta T\\ &= 5\times 0.092\times (80^\circ-20^\circ)\\&= 27.6 \quad {\rm cal} \end{align*} So, it would require 27.6 calories of heat energy to increase the temperature of this substance from 20°C to 80°C.

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Problem (2): How much heat is absorbed by a 20 g granite boulder as energy from the sun causes its temperature to change from 10°C to 29°C? (Specific heat capacity of granite is 0.1 cal/g.°C) 

Solution : to raise the temperature of the granite boulder from 10°C to 29°C, we must add $Q=mc\Delta T$ energy to the granite as below \begin{align*} Q&=mc\Delta T\\ &=20 \times 0.1\times (29^\circ-10^\circ)\\&=38\quad {\rm cal}\end{align*} So, it would require 38 calories of heat energy to increase the temperature of this granite boulder from 10°C to 29°C. This is the amount of energy we must add to the boulder.

In all these example problems, there is no change in the state of the substance. If there were a change in the phase of matter (solid $\Leftrightarrow$ liquid or to liquid$\Leftrightarrow$ gas) read the following page to learn more:

Solved problems on latent heat of fusion

Solved Problems on latent heat of vaporization

Problem (3): How much heat is released when 30 g of water at 96°C cools to 25°C? The specific heat of water is 1 cal/g.°C. 

Solution : the amount of energy released is obtained by formula $Q=mc\Delta T$ as below \begin{align*} Q&=mc\Delta T\\&=30\times 1\times (25^\circ-96^\circ)\\&= -2130\quad {\rm cal}\end{align*} The negative sign in the result indicates that the energy is being released from the water. This is because the temperature of the water is decreasing, which means it is losing heat energy.

Therefore, $2130$ calories of heat energy are released from the water when its temperature decreases from 96°C to 25°C. This energy could be transferred to the surrounding environment or used to do work, depending on the specific circumstances.

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Problem (4):  If a 3.1 g ring is heated using 10.0 calories, its temperature rises 17.9°C. Calculate the specific heat capacity of the ring.  

Solution : Since the given heat causes a change in the temperature of the ring, the amount of heat is obtained by the formula $Q=mc(T_f-T_i)$. By putting known values into it and solving for the unknown value, the specific heat of the ring is calculated as below, \begin{align*} c&=\frac{Q}{m(T_f-T_i)}\\ \\ &=\frac{10}{3.1\times 17.9^\circ}\\ \\&=0.18\quad {\rm cal/g\cdot ^\circ C}\end{align*} So, the specific heat of the ring is calculated to be $0.18\,{\rm cal/g\cdot  ^\circ C}$. This value tells us how much heat is required to raise the temperature of $1$ gram of the ring by $1$ degree Celsius. Note that in this problem, the difference between temperatures is not the initial or final temperatures.

Problem (5): The temperature of a sample of water increases from 20°C to 46.6°C as it absorbs 5650 calories of heat. What is the mass of the sample? (Specific heat of water is 1.0 cal/g.°C)

Solution : As before, using heat formula and solving for mass $m$, we get \begin{align*} m&=\frac{Q}{c\Delta T}\\\\ &=\frac{5650}{1\times (46.6^\circ-20^\circ)}\\ \\&=212.4\quad {\rm g}\end{align*}

Problem (6):  The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat. What is the specific heat of iron?

Solution : In the  specific heat problems , we learned that the specific heat is defined as the amount of heat energy required to change the temperature of a sample with mass $m$ by $\Delta T$.

Here, energy is released by as much as 47 calories, so we must put it with a negative sign into the equation. Thus, we have \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\&=\frac{-47}{10\times (25^\circ-50.4^\circ)}\\ \\&= 0.185\quad {\rm cal/g\cdot ^\circ \! C}\end{align*} 

Problem (7): A 4.50 g coin of copper absorbed 54 calories of heat. What was the final temperature of the copper if the initial temperature was 25°C? The specific heat of copper is 0.092 cal/g.°C.

Solution : Let $T_i$ and $T_f$ be the initial and final temperatures of the copper coin. Again using formula $Q=mc(T_f-T_i)$ and solving for final temperature $T_f$, we have \begin{align*} T_f&=\frac{Q}{mc}+T_i \\ \\ &=\frac{54}{0.092\times 4.5}+25^\circ\\ \\ &=155.43\,{\rm ^\circ C}\end{align*}

Problem (8): A 155 g sample of an unknown substance was heated from 25°C to 40°C. In the process, the substance absorbed 569 calories of energy. What is the specific heat of the substance? 

Solution : In the heat formula $Q=mc\Delta T$, the specific heat of any substance is denoted by $c$. Putting known values into this formula and solving for unknown specific heat, we get \begin{align*} c&=\frac{Q}{m\Delta T}\\ \\ &=\frac{569}{155\times (40^\circ-25^\circ)}\\ \\&=0.244\quad {\rm cal/g\cdot^\circ\! C} \end{align*}

Problem (9): What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its temperature changes from 25°C to 20°C?

Solution : same as above, we have \begin{align*} c&=\frac{Q}{m(T_f-T_i)}\\ \\&=\frac{12}{2.5\times (20^\circ-25^\circ)}\\\\&=0.96\quad {\rm cal/g\cdot ^\circ \! C}\end{align*}

Problem (10): When 3 kg of water is cooled from 80°C to 10°C, how much heat energy is lost? (specific heat of water is $c_W=4.179\,{\rm J/g\cdot ^\circ C}$)

Solution : the heat has led to a change in temperature, so we must use the formula $Q=mc\Delta T$ to find the lost heat as shown below: \begin{align*} Q&=mc(T_f-T_i)\\&=3000\times 4.179\times (10^\circ-80^\circ)\\&=-877590\quad {\rm J} \\ or &=-877.590\quad {\rm kJ}\end{align*} Note that in the above calculation, the value of specific heat is given in grams, and the weight of water is in kilograms. Therefore, first convert them into grams or kilograms and then continue to solve the problem. Here, we converted 3 kg to 3000 g.  

The negative sign indicates that the heat is released from the water.

Problem (11): How much heat is needed to raise a 0.30 kg piece of aluminum from 30°C to 150°C? ($c_{Al}=0.9\,{\rm J/g\cdot ^\circ C}$)

Solution: Let $T_f$ and $T_i$ be the initial and final temperatures of the aluminum so the required heat is computed as below \begin{align*} Q&=mc(T_f-T_i)\\&=0.3\times 900\times (150^\circ-30^\circ)\\&=-32400\quad {\rm J}\\ or &=-32.4\quad {\rm kJ}\end{align*} Here, we converted specific heat in SI units.

Problem (12): Calculate the temperature change when: (a) 10.0 kg of water loses 232 kJ of heat. ($c_W=4.179\,{\rm J/g\cdot ^\circ C}$) (b) 1.96 kJ of heat is added to 500 g of copper.($c_{Cu}=0.385\,{\rm J/g\cdot ^\circ C}$)

Solution : In both parts, we use the heat formula when temperature changes, $Q=mc(T_f-T_i)$.  (a) Substituting known values $m=10\,{\rm kg}$ and $Q=232\,{\rm kJ}$ into the above equation and solving for the change in temperature $\Delta T=T_f-T_i$, we get: \begin{align*} \Delta T&= \frac{Q}{mc}\\ \\&=\frac{-232000}{10\times 4179}\\ \\&=-5.55\,{\rm ^\circ C}\end{align*} Since water loses heat energy (which justifies why we inserted a minus sign for Q), its temperature must be decreasing. In the above, kJ means 1000 J of energy. 

(b) Heat is added to the water, so $Q>0$ must be inserted into the formula, \begin{align*}\Delta T&=\frac{Q}{mc}\\ \\&=\frac{1960}{0.5\times 385}\\ \\&=10.18\,{\rm ^\circ C}\end{align*}

Problem (13): When heated, the temperature of a water sample increased from 15°C to 39°C.  It absorbed 4300 joules of heat.  What is the mass of the sample?

Solution : putting known values into the equation $Q=mc(T_f-T_i)$ and solving for unknown mass, we get \begin{align*} m&=\frac{Q}{c(T_f-T_i)}\\ \\ &=\frac{4300}{4179\times (15^\circ-39^\circ)}\\ \\&=0.0428\quad {\rm kg}\\ \\ or &=42.8\quad {\rm g} \end{align*}

Problem (14): 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu?  

Solution : the necessary energy is calculated as below: \begin{align*} Q&=mc(T_f-T_i)\\&=5\times 0.385\times (20^\circ-80^\circ)\\&=115.5\quad {\rm J}\end{align*} So, the necessary energy or heat absorbed by the object is calculated to be $\rm 115.5\, J$. This value tells us how much heat energy is required to change the temperature of the $\rm 5\, g$ object by $60$ degrees Celsius. 

Problem (15): The temperature of a sample of water increases from 20°C to 46.6°C as it absorbs 5650 Joules of heat. What is the mass of the sample? 

Solution : known values are $T_i={\rm 20^\circ C}$, $T_f={\rm 46.6^\circ C}$ and $Q=5650\,{\rm J}$. We can rearrange the formula to solve for ($m$): \begin{align*} m&=\frac{Q}{c(T_f-T_i)}\\ \\&=\frac{5650}{4179\times (46.6^\circ-20^\circ)}\\ \\ &=0.0508\quad {\rm kg} \\ \\ or &=50.8\quad {\rm g} \end{align*} So, the mass of the water sample is approximately ($50.7$) grams.

Author : Dr. Ali Nemati Page Created: 3/9/2021  

© 2015 All rights reserved. by Physexams.com

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Temperature Worksheets

Thermometers are tools used to measure how hot or cold something is. The temperature worksheets are sure to provide 1st grade through 8th grade students with adequate practice in reading thermometers, shading them, comparing temperatures, ordering them from the warmest to the coldest and vice versa, converting between Celsius and Fahrenheit scales, converting temperatures between Kelvin and Celsius, and to comprehend the increase or decrease in temperature. Drawing a thermometer every time you need to practice a skill is not just time-consuming, it often hinders practice. Our printable thermometer templates remove these barriers. The free temperature worksheets offered here will set you on the right path!

Reading Thermometers Worksheets

Amanda plans snowmobiling with her friends. Her thermometer reads 24°F; just perfect for the activity. Reading a thermometer to tell the temperature is a skill that kids of grade 1, grade 2, grade 3, and grade 4 will become conversant in, if they use these reading thermometer pdfs.

(24 Worksheets)

Comparing Higher and Lower Temperatures

Identifying the thermometer with a higher or a lower temperature is no longer a tough nut to crack. Observe each pair of thermometers in these comparing temperatures worksheets and choose the option that best answers the question.

• Download the set

Comparing Temperatures | Warmer or Colder?

Read the temperature and check the thermometer indicating a lesser value for cooler thermometers and with a higher value for warmer thermometers. Instruct students of grade 5, grade 6, and grade 7 to look closely, and the level of shading will give them a hint.

Ordering Temperatures from Warmest to Coldest

Rich with scads of cool and warm thermometers, these pdfs breathe new life into your ordering temperature practice. Reading the scales and arranging the temperatures from the coldest to the warmest is no big deal.

Ordering Temperatures from Coldest to Warmest

Continue to allow your thermometer-reading skill to feed on enormous practice! Students of 6th grade, 7th grade, and 8th grade arrange temperatures from the hottest to the coldest in these temperature worksheets keeping a watch on the tricky negative temperatures!

Arranging Temperatures in Ascending and Descending Order

Top up practice with practice! Whether it is arranging temperatures in ascending or descending order, children do it effortlessly working their way through these printable ordering thermometer worksheets that transform them into champs!

Converting between Celsius and Fahrenheit Worksheets

Did you know 0°C = 32°F and -40°C = -40°F? With different units being used in different countries, it becomes essential for 5th grade through 8th grade kids to learn the formulas to convert temperatures between the two units °F and °C.

(15 Worksheets)

Converting between Celsius and Kelvin Worksheets

The Kelvin scale is used to measure extremely cold or hot temperatures. Switch temperatures between °C and K easily, by adding 273.15 to a Celsius temperature to change it to Kelvin and subtract 273.15 to do vice versa in these Kelvin and Celsius conversion pdfs.

(12 Worksheets)

Adding and Subtracting Temperatures

Temperature has a reputation for fickleness! Grade 6, grade 7, and grade 8 learners read the thermometers, add temperatures when there is an increase or rise and subtract when there is a fall or decrease. Be vigilant with the negative temperatures!

Printable Blank Thermometer Templates

Whether it is a single blank thermometer template or a collection of 8 thermometers in one printable template, with temperature readings between 0 and 50, 0 and 100, or -50 and 50 degrees in both Celsius and Fahrenheit scales, we have them all pre-made to save your time!

(8 Worksheets)

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» Metric Unit Conversion

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Fahrenheit to Celsius Exercises

Converting fahrenheit to celsius practice problems.

It’s time to sharpen those temperature conversion skills! For this activity, we’ll focus on how to convert temperatures from Fahrenheit to Celsius . Enjoy!

Here’s the Fahrenheit to Celsius conversion formula for your reference.

Problem 1 : Convert 77 degrees Fahrenheit to Celsius.

25 degrees Celsius or 25° C

Problem 2 : Convert – 32.8° F to °C.

– 36° C

Problem 3 : What is 113 degrees Fahrenheit in Celsius?

45 degrees Celsius or 45° C

Problem 4 : Convert – 10.3° F to °C.

– 23.5° C

Problem 5 : What is – 40° F in Celsius?

– 40° C

Problem 6 : Mom preheated the oven at 450° F. She then increased the temperature by 5° F to roast the chicken for 10 minutes. What was the roasting temperature in Celsius?

Problem 7 : Convert 194° F to °C.

Problem 8 : What is 0° F in Celsius? Round your answer to the nearest hundredth.

About – 17.78° C or ≈ – 17.78° C

Problem 9 : It is hot outside. The temperature is at 95° F. What is it in Celsius?

Problem 10 : Convert 554° F to °C

You might also like these tutorials:

• Converting Fahrenheit to Celsius
• Converting Celsius to Fahrenheit

9.5 Measuring Temperature

Learning objectives.

After completing this section, you should be able to:

• Convert between Fahrenheit and Celsius.
• Identify reasonable values for temperature applications.
• Solve application problems involving temperature.

When you touch something and it feels warm or cold, what is that really telling you about that substance? Temperature is a measure of how fast atoms and molecules are moving in a substance, whether that be the air, a stove top, or an ice cube. The faster those atoms and molecules move, the higher the temperature.

In the metric system, temperature is measured using the Celsius (°C) scale. Because temperature is a condition of the physical properties of a substance, the Celsius scale was created with 100 degrees separating the point at which water freezes, 0 °C, and the point at which water boils, 100 °C. Scientifically, these are the points at which water molecules change from one state of matter to another—from solid (ice) to liquid (water) to gas (water vapor).

When reading temperatures, it’s important to look beyond the degree symbol to determine which temperature scale the units express. For example, 13 °C reads “13 degrees Celsius,” indicating that the temperature is expressed using the Celsius scale, while 13 °F reads “13 degrees Fahrenheit,” indicating that the temperature is expressed using the Fahrenheit scale.

Misconceptions About Temperature

How Many Temperature Scales Are There?

Did you know that in addition to Fahrenheit and Celsius, there is a third temperature scale widely used throughout the world? The Kelvin scale starts at absolute zero, the lowest possible temperature at which there is no heat energy present at all. It is primarily used by scientists to measure very high or very low temperatures when water is not involved.

Converting Between Fahrenheit and Celsius Temperatures

Understanding how to convert between Fahrenheit and Celsius temperatures is an essential skill in understanding metric temperatures. You likely know that below 32 °F means freezing temperatures and perhaps that the same holds true for 0 °C. While it may be difficult to recall that water boils at 212 °F, knowing that it boils at 100 °C is a fairly easy thing to remember.

But what about all the temperatures in between? What is the temperature in degrees Celsisus on a scorching summer day? What about a cool autumn afternoon? If a recipe instructs you to preheat the oven to 350 °F, what Celsius temperature do you set the oven at?

Figure 9.15 lists common temperatures on both scales, because we don’t use Celsius temperatures daily it’s difficult to remember them. Fortunately, we don’t have to. Instead, we can convert temperatures from Fahrenheit to Celsius and from Celsius to Fahrenheit using a simple algebraic expression.

The formulas used to convert temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in Table 9.3 .

Example 9.40

Converting temperatures from fahrenheit to celsius.

A recipe calls for the oven to be set to 392 °F. What is the temperature in Celsius?

Use the formula in Table 9.3 to convert from Fahrenheit to Celsius.

So, 392 °F is equivalent to 200 °C.

Your Turn 9.40

Example 9.41, converting temperatures from celsius to fahrenheit.

On a sunny afternoon in May, the temperature in London was 20 °C. What was the temperature in degrees Fahrenheit?

Use the formula in Table 9.3 to convert from Celsius to Fahrenheit.

The temperature was 68 °F.

Your Turn 9.41

Example 9.42, comparing temperatures in celsius and fahrenheit.

A manufacturer requires a vaccine to be stored in a refrigerator at temperatures between 36 °F and 46 °F. The refrigerator in the local pharmacy cools to 3 °C. Can the vaccine be stored safely in the pharmacy’s refrigerator?

Then, compare the temperatures.

Yes. 37.4 °F falls within the acceptable range to store the vaccine, so it can be stored safely in the pharmacy’s refrigerator.

Your Turn 9.42

Reasonable values for temperature.

While knowing the exact temperature is important in most cases, sometimes an approximation will do. When trying to assess the reasonableness of values for temperature, there is a quicker way to convert temperatures for an approximation using mental math. These simpler formulas are listed in Table 9.4 .

The formulas used to estimate temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in Table 9.4 .

Temperature Conversion Trick

Example 9.43

Using benchmark temperatures to determine reasonable values for temperatures.

Which is the more reasonable value for the temperature of a freezer?

We know that water freezes at 0 °C. So, the more reasonable value for the temperature of a freezer is −5 °C, which is below 0 °C. At temperature of 5 °C is above freezing.

Your Turn 9.43

Example 9.44, using estimation to determine reasonable values for temperatures.

The average body temperature is generally accepted as 98.6 °F. What is a reasonable value for the average body temperature in degrees Celsius:

• 64.3 °C, or

To estimate the average body temperature in degrees Celsius, subtract 30 from the temperature in degrees Fahrenheit, and divide the result by 2.

A reasonable value for average body temperature is 34.3 °C.

Your Turn 9.44

Example 9.45, using conversion to determine reasonable values for temperatures.

Which is a reasonable temperature for storing chocolate:

Use the formula in Table 9.3 to determine the temperature in degrees Fahrenheit.

A temperature of 82.4 °F would be too hot, causing the chocolate to melt. A temperature of 35.6 °F is very close to freezing, which would affect the look and feel of the chocolate. So, a reasonable temperature for storing chocolate is 18 °C, or 64.4 °F.

Your Turn 9.45

Solving application problems involving temperature.

Whether traveling abroad or working in a clinical laboratory, knowing how to solve problems involving temperature is an important skill to have. Many food labels express sizes in both ounces and grams. Most rulers and tape measures are two-sided with one side marked in inches and feet and the other in centimeters and meters. And while many thermometers have both Fahrenheit and Celsius scales, it really isn’t practical to pull out a thermometer when cooking a recipe that uses metric units. Let’s review at few instances where knowing how to fluently use the Celsius scale helps solve problems.

Example 9.46

Using subtraction to solve temperature problems.

The temperature in the refrigerator is 4 °C. The temperature in the freezer is 21 °C lower. What is the temperature in the freezer?

Use subtraction to find the difference.

So, the temperature in the freezer is −17 °C.

Your Turn 9.46

Example 9.47, using addition to solve temperature problems.

A scientist was using a liquid that was 35 °C. They needed to heat the liquid to raise the temperature by 6 °C. What was the temperature after the scientist heated it?

Use addition to find the new temperature.

The temperature of the liquid was 41 °C after the scientist heated it.

Your Turn 9.47

Example 9.48, solving complex temperature problems.

The optimum temperature for a chemical compound to develop its unique properties is 392 °F. When the heating process begins, the temperature of the compound is 20 °C. For safety purposes the compound can only be heated 9 °C every 15 minutes. How long until the compound reaches its optimum temperature?

Step 1: Determine the optimum temperature in degrees Celsius using the formula in Table 9.3 .

Step 2: Subtract the starting temperature.

Step 3: Determine the number of 15-minute cycles needed to heat the compound to its optimum temperature.

Step 4: Multiply the number of cycles needed by 15 minutes and convert the product to hours and minutes.

So, it will take 2 hours and 15 minutes for the compound to reach its optimum temperature.

Your Turn 9.48

Learn the Metric System in 5 Minutes

Check Your Understanding

Section 9.5 exercises.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

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• Publication date: Mar 22, 2023
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• Book URL: https://openstax.org/books/contemporary-mathematics/pages/1-introduction
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9.6: Measuring Temperature

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Learning Objectives

After completing this section, you should be able to:

• Convert between Fahrenheit and Celsius.
• Identify reasonable values for temperature applications.
• Solve application problems involving temperature.

When you touch something and it feels warm or cold, what is that really telling you about that substance? Temperature is a measure of how fast atoms and molecules are moving in a substance, whether that be the air, a stove top, or an ice cube. The faster those atoms and molecules move, the higher the temperature.

In the metric system, temperature is measured using the Celsius (°C) scale. Because temperature is a condition of the physical properties of a substance, the Celsius scale was created with 100 degrees separating the point at which water freezes, 0 °C, and the point at which water boils, 100 °C. Scientifically, these are the points at which water molecules change from one state of matter to another—from solid (ice) to liquid (water) to gas (water vapor).

When reading temperatures, it’s important to look beyond the degree symbol to determine which temperature scale the units express. For example, 13 °C reads “13 degrees Celsius,” indicating that the temperature is expressed using the Celsius scale, while 13 °F reads “13 degrees Fahrenheit,” indicating that the temperature is expressed using the Fahrenheit scale.

Misconceptions About Temperature

How Many Temperature Scales Are There?

Did you know that in addition to Fahrenheit and Celsius, there is a third temperature scale widely used throughout the world? The Kelvin scale starts at absolute zero, the lowest possible temperature at which there is no heat energy present at all. It is primarily used by scientists to measure very high or very low temperatures when water is not involved.

Converting Between Fahrenheit and Celsius Temperatures

Understanding how to convert between Fahrenheit and Celsius temperatures is an essential skill in understanding metric temperatures. You likely know that below 32 °F means freezing temperatures and perhaps that the same holds true for 0 °C. While it may be difficult to recall that water boils at 212 °F, knowing that it boils at 100 °C is a fairly easy thing to remember.

But what about all the temperatures in between? What is the temperature in degrees Celsisus on a scorching summer day? What about a cool autumn afternoon? If a recipe instructs you to preheat the oven to 350 °F, what Celsius temperature do you set the oven at?

Figure 9.16 lists common temperatures on both scales, because we don’t use Celsius temperatures daily it’s difficult to remember them. Fortunately, we don’t have to. Instead, we can convert temperatures from Fahrenheit to Celsius and from Celsius to Fahrenheit using a simple algebraic expression.

The formulas used to convert temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in Table 9.3.

Example 9.40

Converting temperatures from fahrenheit to celsius.

A recipe calls for the oven to be set to 392 °F. What is the temperature in Celsius?

Use the formula in Table 9.3 to convert from Fahrenheit to Celsius.

C = 5 9 ( F − 32 ) C = 5 9 ( 392 − 32 ) C = 5 9 ( 360 ) C = 200 C = 5 9 ( F − 32 ) C = 5 9 ( 392 − 32 ) C = 5 9 ( 360 ) C = 200

So, 392 °F is equivalent to 200 °C.

Your Turn 9.40

Example 9.41, converting temperatures from celsius to fahrenheit.

On a sunny afternoon in May, the temperature in London was 20 °C. What was the temperature in degrees Fahrenheit?

Use the formula in Table 9.3 to convert from Celsius to Fahrenheit.

F = 9 5 C + 32 F = 9 5 ( 20 ) + 32 F = 36 + 32 F = 68 F = 9 5 C + 32 F = 9 5 ( 20 ) + 32 F = 36 + 32 F = 68

The temperature was 68 °F.

Your Turn 9.41

Example 9.42, comparing temperatures in celsius and fahrenheit.

A manufacturer requires a vaccine to be stored in a refrigerator at temperatures between 36 °F and 46 °F. The refrigerator in the local pharmacy cools to 3 °C. Can the vaccine be stored safely in the pharmacy’s refrigerator?

F = 9 5 C + 32 F = 9 5 ( 3 ) + 32 F = 5.4 + 32 F = 37.4 F = 9 5 C + 32 F = 9 5 ( 3 ) + 32 F = 5.4 + 32 F = 37.4

Then, compare the temperatures.

36 F ° < 37.4 F ° < 46 F ° 36 F ° < 37.4 F ° < 46 F °

Yes. 37.4 °F falls within the acceptable range to store the vaccine, so it can be stored safely in the pharmacy’s refrigerator.

Your Turn 9.42

Reasonable values for temperature.

While knowing the exact temperature is important in most cases, sometimes an approximation will do. When trying to assess the reasonableness of values for temperature, there is a quicker way to convert temperatures for an approximation using mental math. These simpler formulas are listed in Table 9.4.

The formulas used to estimate temperatures from Fahrenheit to Celsius or from Celsius to Fahrenheit are outlined in Table 9.4.

Temperature Conversion Trick

Example 9.43

Using benchmark temperatures to determine reasonable values for temperatures.

Which is the more reasonable value for the temperature of a freezer?

• 5 °C or
• –5 °C?

We know that water freezes at 0 °C. So, the more reasonable value for the temperature of a freezer is −5 °C, which is below 0 °C. At temperature of 5 °C is above freezing.

Your Turn 9.43

• 48 °C or
• 148 °C?

Example 9.44

Using estimation to determine reasonable values for temperatures.

The average body temperature is generally accepted as 98.6 °F. What is a reasonable value for the average body temperature in degrees Celsius:

• 98.6 °C,
• 64.3 °C, or
• 34.3 °C?

To estimate the average body temperature in degrees Celsius, subtract 30 from the temperature in degrees Fahrenheit, and divide the result by 2.

( 98.6 − 30 ) 2 = 68.6 2 = 34.3 ( 98.6 − 30 ) 2 = 68.6 2 = 34.3

A reasonable value for average body temperature is 34.3 °C.

Your Turn 9.44

• 102 °C,
• 37 °C, or

Example 9.45

Using conversion to determine reasonable values for temperatures.

Which is a reasonable temperature for storing chocolate:

• 18 °C, or

Use the formula in Table 9.3 to determine the temperature in degrees Fahrenheit.

F = 9 5 C + 32 F = 9 5 ( 28 ) + 32 = 82.4 F = 9 5 ( 18 ) + 32 = 64.4 F = 9 5 ( 2 ) + 32 = 35.6 F = 9 5 C + 32 F = 9 5 ( 28 ) + 32 = 82.4 F = 9 5 ( 18 ) + 32 = 64.4 F = 9 5 ( 2 ) + 32 = 35.6

A temperature of 82.4 °F would be too hot, causing the chocolate to melt. A temperature of 35.6 °F is very close to freezing, which would affect the look and feel of the chocolate. So, a reasonable temperature for storing chocolate is 18 °C, or 64.4 °F.

Your Turn 9.45

• 240 °C,
• 60 °C, or

Solving Application Problems Involving Temperature

Whether traveling abroad or working in a clinical laboratory, knowing how to solve problems involving temperature is an important skill to have. Many food labels express sizes in both ounces and grams. Most rulers and tape measures are two-sided with one side marked in inches and feet and the other in centimeters and meters. And while many thermometers have both Fahrenheit and Celsius scales, it really isn’t practical to pull out a thermometer when cooking a recipe that uses metric units. Let’s review at few instances where knowing how to fluently use the Celsius scale helps solve problems.

Example 9.46

Using subtraction to solve temperature problems.

The temperature in the refrigerator is 4 °C. The temperature in the freezer is 21 °C lower. What is the temperature in the freezer?

Use subtraction to find the difference.

4 − 21 = − 17 4 − 21 = − 17

So, the temperature in the freezer is −17 °C.

Your Turn 9.46

Example 9.47, using addition to solve temperature problems.

A scientist was using a liquid that was 35 °C. They needed to heat the liquid to raise the temperature by 6 °C. What was the temperature after the scientist heated it?

Use addition to find the new temperature.

35 + 6 = 41 35 + 6 = 41

The temperature of the liquid was 41 °C after the scientist heated it.

Your Turn 9.47

Example 9.48, solving complex temperature problems.

The optimum temperature for a chemical compound to develop its unique properties is 392 °F. When the heating process begins, the temperature of the compound is 20 °C. For safety purposes the compound can only be heated 9 °C every 15 minutes. How long until the compound reaches its optimum temperature?

Step 1: Determine the optimum temperature in degrees Celsius using the formula in Table 9.3.

Step 2: Subtract the starting temperature.

200 C ° − 20 C ° = 180 C ° 200 C ° − 20 C ° = 180 C °

Step 3: Determine the number of 15-minute cycles needed to heat the compound to its optimum temperature.

180 ÷ 9 = 20 180 ÷ 9 = 20

Step 4: Multiply the number of cycles needed by 15 minutes and convert the product to hours and minutes.

15 ⁢ minutes × 9 = 135 ⁢ minutes 135 ⁢ minutes = 2 ⁢ hours ⁢ 15 ⁢ minutes 15 ⁢ minutes × 9 = 135 ⁢ minutes 135 ⁢ minutes = 2 ⁢ hours ⁢ 15 ⁢ minutes

So, it will take 2 hours and 15 minutes for the compound to reach its optimum temperature.

Your Turn 9.48

Learn the Metric System in 5 Minutes

Check Your Understanding

Section 9.5 exercises.

• 40 °C, or

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Chemistry: Gas Laws

Temperature Word Problems

Wendy Rolle

Word Problems involving Temperature

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