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Temperature is how hot something is; internal energy is the total energy of its particles; warming takes $E = mc\Delta T$.
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By the end of this lesson you will be able to tell temperature from internal energy and find the energy a temperature rise needs.
You know that everything is made of tiny particles that are always moving, and that energy can be transferred and measured in joules. You can read a thermometer in degrees Celsius. This lesson separates two ideas that are often confused, temperature and energy, and puts numbers on heating.
| Term | What it means |
|---|---|
| Temperature | How hot something is: the average energy of its particles, in degrees Celsius. |
| Internal energy | The total energy of all the particles in an object. |
| Heat | Energy flowing from a hotter object to a colder one. |
| Specific heat capacity | The energy to warm one kilogram by one degree, $c$, in J/(kg·°C). |
| Temperature rise | The change in temperature, $\Delta T$, final minus starting. |
| Thermal equilibrium | When two objects in contact reach the same temperature. |
Every object is made of particles that jiggle and move. Two ideas describe their energy:
The energy needed to warm a material depends on its mass, what it is made of, and how much hotter it gets:
$$E = mc\Delta T,$$
where $c$ is the specific heat capacity. Water's is $4200$ J/(kg·°C), one of the highest of any common material.
Another way: picture
Picture a spark from a sparkler landing on your hand. The spark is over a thousand degrees, yet it barely stings, because it is so tiny that it carries almost no energy. A cup of hot cocoa at seventy degrees can burn you badly, because it has far more mass and so far more energy to give.
Another way: steps
All matter is made of particles, atoms and molecules, that are always moving. In a solid they vibrate in place; in a liquid they slide past each other; in a gas they fly about freely. The hotter the material, the faster they move.
Temperature measures how fast, on average, the particles are moving. A thermometer reads that average, whether it is dipped in a thimble of water or a swimming pool.
Internal energy is the total energy of all the particles together. A swimming pool at twenty-five degrees has an enormous internal energy, because it contains so many particles, even though each is only moving gently.
A cup of boiling water is hotter, with faster particles, but there are so few of them that its total internal energy is far smaller than the pool's.
When two objects at different temperatures touch, energy flows from the hotter one to the colder one. That flow of energy is called heat. It continues until both reach the same temperature.
Heat always flows from hot to cold, never the other way on its own. A cold drink warms up on a summer day, and a hot bowl of soup cools down, until each matches the air around it.
Different materials need different amounts of energy to warm up. It takes $4200$ joules to warm one kilogram of water by one degree, but only about $900$ to warm aluminum and $450$ to warm iron by the same amount.
This number, the specific heat capacity, tells you how stubborn a material is about changing temperature. Water is very stubborn: it takes a lot of energy to heat and releases a lot as it cools.
The energy needed to warm something is its mass, times its specific heat capacity, times its temperature rise: $E = mc\Delta T$. Warming two kilograms of water by thirty degrees takes $2 \times 4200 \times 30 = 252000$ joules.
Each factor makes sense on its own. More material needs more energy; a stubborn material needs more energy; a bigger temperature change needs more energy.
Rearranging the equation gives the temperature rise from a known energy: $\Delta T = E/(mc)$. The same energy warms a small mass a lot and a large mass only a little.
It also warms materials with low specific heat capacity much more. A cast-iron skillet heats up quickly on the stove, while a pot of water takes much longer to reach the same temperature.
An electric heater supplies energy at a steady rate, its power. Dividing the energy needed by the power gives the time. A two-thousand-watt kettle needs about four minutes to bring a liter and a half of water to a boil.
Real heating takes a little longer, because some energy escapes into the kettle, the air and the counter. The equation gives the best case, with no losses.
Checking an answer. Heating water takes thousands of joules per kilogram per degree, so everyday answers run to tens or hundreds of thousands of joules. Twice the mass must take twice the energy.
Experiments show that the energy needed to warm a material is proportional to its mass and to the temperature rise. The constant linking them depends only on the material, which is why it can be measured once and listed in tables.
Energy is conserved, so the energy a heater supplies, minus any lost, all goes into the water's internal energy, raising its temperature.
Water's large specific heat capacity shapes weather and climate. Oceans and large lakes warm slowly in summer and cool slowly in winter, which keeps coastal cities milder than inland ones.
San Francisco, beside the Pacific, rarely gets very hot or very cold, while Kansas City, far from any ocean, swings from hot summers to cold winters. The Great Lakes similarly soften the climate of the cities along their shores.
Scientists and most of the world measure temperature in degrees Celsius, where water freezes at zero and boils at one hundred at sea level. American weather forecasts use degrees Fahrenheit, where water freezes at thirty-two and boils at two hundred twelve.
A change of one degree Celsius equals a change of one and eight tenths degrees Fahrenheit. The heating equation uses Celsius, so convert Fahrenheit first.
The most common error is using the final temperature instead of the temperature rise. Heating water from twenty to seventy degrees is a rise of fifty, not seventy.
Another is confusing temperature with internal energy, thinking the hottest object must have the most energy. A large, lukewarm object can hold far more energy than a small, very hot one.
A few benchmarks help. Warming a cup of water for tea takes about a hundred thousand joules. Heating a bathtub of water takes about ten million. Heating a home's water heater tank from cold takes about twenty-five million.
Comparing an answer with these helps catch a missing factor, such as a forgotten mass or specific heat capacity.
The same equation works in reverse. As an object cools, it gives off energy equal to its mass times its specific heat capacity times its temperature drop. A hot water bottle stays warm for hours because water gives off a lot of energy as it cools.
That is also why a hot stone or brick was once used to warm beds on cold nights: it stored energy from the fire and released it slowly.
Heat flows from hot to cold, so keeping something hot means slowing that flow. Insulation, such as the foam in a cooler or the fiberglass in an attic, traps air and slows the flow of heat.
A well-insulated home in Minnesota keeps its heat longer in winter, using less energy for heating. A thermos keeps soup hot for hours by surrounding it with a vacuum, which heat crosses only very slowly.
In a lab, the specific heat capacity of a metal can be found by heating a block with an electric heater of known power for a measured time, and recording the temperature rise.
The energy supplied is power times time. Dividing by the block's mass and its temperature rise gives the specific heat capacity, which can be compared with a table to identify the metal.
On a sunny day at the beach, the sand can burn your feet while the ocean water feels cool. Both receive about the same sunlight, but sand has a much lower specific heat capacity than water.
The same energy warms the sand several times as much. At night the sand cools quickly too, while the water stays almost the same temperature, which is why ocean breezes change direction between day and night.
Most American homes have a water heater, often a forty-gallon tank holding about one hundred fifty kilograms of water. It heats water from the temperature it arrives at from the city pipes up to about fifty degrees Celsius.
The energy needed depends on the climate. In Minneapolis, winter tap water can arrive at about seven degrees, so each tank needs a rise of over forty degrees, about twenty-seven megajoules. In Miami, where the water arrives near twenty-seven degrees, the rise is only about twenty-three, roughly half the energy. That is one reason water heating costs more in northern states, and why solar water heaters are especially popular in the sunny South.
The Great Lakes hold an enormous mass of water, with a very high specific heat capacity. In autumn they cool slowly, staying warmer than the cold air that sweeps down from Canada.
As freezing air crosses the warmer lakes, it picks up heat and moisture, then drops it as heavy snow on the downwind shores. Cities like Buffalo, New York, and parts of Michigan get famous lake-effect snowstorms, sometimes over a meter in a day. The lakes' stubbornness about changing temperature, their high heat capacity, is what keeps them warm enough to feed these storms well into winter.
In everyday speech, heat and temperature are often used as if they meant the same thing. But temperature measures how hot something is, the average energy of its particles, while internal energy is the total. A large tub of warm water holds far more energy than a small cup of boiling water.
A related error is to use the final temperature in $E = mc\Delta T$ instead of the temperature rise. Heating water from twenty to seventy degrees is a rise of fifty degrees.
A pot holds $3$ kg of water at $20$ °C, to be heated to $100$ °C. Find the temperature rise.
$\Delta T = 100 - 20 = 80\ \text{°C}$
Final minus starting.
Write the heating formula.
$E = mc\Delta T$
Mass, capacity, rise.
Substitute the values.
$E = 3 \times 4200 \times 80$
Water's capacity.
Evaluate the energy.
$E = 1008000\ \text{J}$
About a million joules.
Find the time on a $2000$ W burner.
$t = \dfrac{1008000}{2000} = 504\ \text{s}$
About eight and a half minutes.
A $2$ kg iron skillet absorbs $36000$ J. Iron's capacity is $450$ J/(kg·°C). Write the rise formula.
$\Delta T = \dfrac{E}{mc}$
Rearranged.
Find the skillet's rise.
$\Delta T = \dfrac{36000}{2 \times 450} = 40\ \text{°C}$
Warms quickly.
Find the rise for $2$ kg of water with the same energy.
$\Delta T = \dfrac{36000}{2 \times 4200} = 4.3\ \text{°C}$
Much less.
Compare the two rises.
$\dfrac{40}{4.3} \approx 9$
About nine times.
Explain the difference.
$\text{water's capacity is about nine times iron's}$
Specific heat capacity.
Say which cools faster off the stove.
$\text{the skillet}$
Less energy stored per degree.
A small pool holds $10000$ kg of water at $25$ °C. Find the energy to warm it by $1$ °C.
$E = 10000 \times 4200 \times 1 = 42000000\ \text{J}$
Forty-two megajoules.
A cup holds $0.25$ kg of water. Find the energy to warm it by $1$ °C.
$E = 0.25 \times 4200 = 1050\ \text{J}$
Tiny by comparison.
Find the energy to heat the cup from $25$ °C to $100$ °C.
$E = 0.25 \times 4200 \times 75 = 78750\ \text{J}$
Under a hundred thousand.
Compare with one degree for the pool.
$78750 \ll 42000000$
The pool needs far more.
Say which has the higher temperature.
$\text{the boiling cup}$
Faster particles.
Say which has more internal energy.
$\text{the pool}$
Far more particles.
State the lesson.
$\text{temperature and energy differ}$
Average versus total.
Write the heating formula.
$E = mc\Delta T$
Mass, capacity, rise.
Substitute the values.
$E = 0.5 \times 4200 \times 20$
In joules.
Evaluate the energy.
How much energy, in J, does it take to warm $0.25$ kg of water by $60$ °C? Water's specific heat capacity is $4200$ J per kilogram per degree Celsius.
Complete the worked solution: an electric kettle of $1000$ W heats $0.5$ kg of water by $60$ °C. With $c = 4200$ J/(kg·°C) and no heat lost, find the energy needed in J, the time in s, and the time in minutes.
Find the energy.
$E = mc\Delta T =$ e
Mass, capacity, rise.
Find the time in seconds.
$t = \dfrac{E}{P} =$ s
Energy over power.
Convert to minutes.
$t \div 60 =$ n
Sixty seconds per minute.
Explain a real kettle.
$\text{takes a little longer}$
Some heat escapes.
Match each idea to its meaning.
| the average energy of an object's particles | the total energy of all an object's particles | the energy to warm one kilogram by one degree | energy flowing from a hotter object to a colder one | |
|---|---|---|---|---|
| temperature | ||||
| internal energy | ||||
| specific heat capacity | ||||
| heat |
A $2000$ W electric burner heats $3$ kg of water by $20$ °C, with all its energy going into the water. Using $c = 4200$ J/(kg·°C), fill in the energy needed in J, the time it takes in s, and the energy needed in J for twice as much water.
| value | |
|---|---|
| energy needed (J) | |
| heating time (s) | |
| energy for twice the water (J) |
A pot holds $1.5$ kg of water, with $c = 4200$ J/(kg·°C). Write the energy needed, in J, as a function of the temperature rise $T$ in degrees Celsius.
Answer:
an aluminum pan has a mass of $0.5$ kg and a specific heat capacity of $900$ J/(kg·°C). If it absorbs $18000$ J of energy, by how many degrees Celsius does its temperature rise?
Answer: °C
A 40-gallon home water heater holds about $151$ kg of water and heats it to $50$ °C. In Miami, Florida, the water comes in from the city pipes at about $27$ °C. With $c = 4200$ J/(kg·°C), how much energy does one full tank of heating take, in MJ?
Answer: MJ
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A pot holds $0.25$ kg of water, with $c = 4200$ J/(kg·°C). Write the energy needed, in J, as a function of the temperature rise $T$ in degrees Celsius.
Answer:
You can use the heating equation. Explain to someone why a bathtub of warm water holds more energy than a cup of boiling water.
28. Your turn: how much energy warms $0.5$ kg of water by $20$ °C, with $c = 4200$ J/(kg·°C)?, step 3
$E = 42000\ \text{J}$
Forty-two kilojoules.