D. Energy and Society · Thermal and Nuclear EnergySpecific Heat Capacity and Q = mcΔT
D3.3 D2.10 D2.1 A1.13About 70 minutes
Learning goals- I can define specific heat capacity and heat capacity and state their units.
- I can explain why water warms and cools slowly and how that affects Canadian weather and climate.
- I can use Q = mcΔT to calculate heat, mass, specific heat capacity, temperature change or final temperature.
- I can use a sign convention for heat gained and heat lost, and report answers with correct units and significant digits.
Words to know
specific heat capacity (c)
The amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C. Unit: J/(kg·°C).
heat capacity
The energy needed to raise the temperature of a whole object by 1 °C. For one material, heat capacity = mc. Unit: J/°C.
quantity of heat (Q)
The thermal energy transferred to or from a substance, in joules. Q = mcΔT when there is no change of state.
temperature change (ΔT)
Final temperature minus initial temperature: ΔT = T₂ − T₁. Positive when warming, negative when cooling.
sign convention
In this unit Q is positive when a substance gains heat and negative when it loses heat.
The sand is hot, the lake is not
On a sunny July afternoon on the shore of Lake Huron, the sand can be too hot to walk on, while the water still feels chilly. Both received about the same sunshine all day. At night, the sand quickly cools down, but the lake is still warm for a late swim.
Water is unusually stubborn about changing temperature. It takes a lot of energy to warm it up and it gives back a lot of energy as it cools. Physicists measure this stubbornness with a property called specific heat capacity.
What specific heat capacity means
The specific heat capacity, c, of a substance is the amount of energy needed to raise the temperature of 1 kg of that substance by 1 °C.
Its unit is joules per kilogram per degree Celsius, written J/(kg·°C). (Because a change of 1 °C equals a change of 1 K, you will also see J/(kg·K); the numbers are the same.)
For water, c = 4.18 × 10³ J/(kg·°C). That means 4180 J of energy warms 1 kg (1 L) of water by just 1 °C. For copper, c is about 390 J/(kg·°C): the same 4180 J would warm 1 kg of copper by more than 10 °C!
Specific heat capacity is a property of the substance, like density. It does not depend on how much you have. The heat capacity of a whole object, by contrast, depends on its mass: heat capacity = mc. A full bathtub has a far bigger heat capacity than a mug of the same water.
| Substance | c (J/(kg·°C)) |
|---|
| water | 4 180 |
| ice | 2 100 |
| steam | 2 000 |
| aluminium | 900 |
| glass | 840 |
| iron | 450 |
| copper | 390 |
| lead | 130 |
| air (dry) | about 1000 |
The equation Q = mcΔT
The energy needed to change the temperature of a substance depends on three things:
- the mass (twice the mass needs twice the energy),
- the substance (its specific heat capacity), and
- the temperature change (twice the change needs twice the energy).
Putting these together:
Q = mcΔT
- Q = quantity of heat (J)
- m = mass (kg)
- c = specific heat capacity (J/(kg·°C))
- ΔT = T₂ − T₁ = final temperature − initial temperature (°C)
Sign convention: if the substance warms up, ΔT is positive, so Q is positive: the substance gained heat. If it cools down, ΔT is negative, so Q is negative: it lost heat.
This equation works only while the substance stays in the same state. During melting or boiling the temperature does not change, so you need a different equation. That is the lesson Changes of State and Latent Heat.
You can rearrange it to find any variable: m = Q/(cΔT), c = Q/(mΔT), ΔT = Q/(mc).
Units and significant digits
Always convert grams to kilograms first (divide by 1000), because c is given per kilogram. 1 L of water has a mass of about 1 kg. Keep all digits until the end, then round your final answer to the fewest significant digits in the data you were given. Large answers are often clearer in kilojoules: 1 kJ = 1000 J.
Worked exampleExample 1: Heating water for tea
How much heat is needed to warm 2.0 kg of water from 15 °C to 100 °C?
- Given: m = 2.0 kg, c = 4.18 × 10³ J/(kg·°C), T₁ = 15 °C, T₂ = 100 °C.
- ΔT = T₂ − T₁ = 100 − 15 = 85 °C.
- Q = mcΔT = (2.0 kg)(4180 J/(kg·°C))(85 °C) = 710 600 J.
- Round to 2 significant digits (2.0 kg and 85 °C have two): Q = 7.1 × 10⁵ J, or 710 kJ. Q is positive because the water gains heat.
Q = 7.1 × 10⁵ J (about 710 kJ) of heat must be added.
Worked exampleExample 2: A cooling frying pan
A 500 g aluminium frying pan cools from 180 °C to 20 °C. How much heat does it lose?
- Convert: m = 500 g = 0.500 kg. c(aluminium) = 900 J/(kg·°C).
- ΔT = 20 − 180 = −160 °C (negative: it cools).
- Q = mcΔT = (0.500)(900)(−160) = −72 000 J.
- The negative sign means heat left the pan. Q = −7.2 × 10⁴ J, so the pan lost 72 kJ to the stove top and air.
Q = −7.2 × 10⁴ J; the pan loses 72 kJ of heat.
The same energy produces very different temperature changes. Since ΔT = Q/(mc), the substance with the smaller specific heat capacity heats up more.Worked exampleExample 3: Identify the metal
A 0.25 kg block of an unknown metal absorbs 2925 J of heat and its temperature rises from 20.0 °C to 50.0 °C. Find its specific heat capacity and identify the metal using the table.
- Rearrange: c = Q/(mΔT).
- ΔT = 50.0 − 20.0 = 30.0 °C.
- c = 2925 J ÷ (0.25 kg × 30.0 °C) = 2925 ÷ 7.5 = 390 J/(kg·°C).
- This matches copper in the table.
c = 3.9 × 10² J/(kg·°C), so the metal is probably copper.
Worked exampleExample 4: Find the final temperature
1.5 kg of water at 12 °C absorbs 2.0 × 10⁵ J of energy from a solar water heater. What is its final temperature?
- ΔT = Q/(mc) = 200 000 ÷ (1.5 × 4180) = 200 000 ÷ 6270 = 31.9 °C.
- T₂ = T₁ + ΔT = 12 + 31.9 = 43.9 °C.
- Round to 2 significant digits: 44 °C.
The water ends at about 44 °C.
Water, lakes and Canadian climate
Because water has such a large specific heat capacity, large bodies of water warm slowly in spring and summer and cool slowly in fall and winter. They act like giant thermal batteries.
- Places near the Great Lakes and the oceans have milder winters and cooler summers than places far inland at the same latitude. Toronto's winters are generally milder than Winnipeg's, partly because the Prairies are far from any large body of water.
- In late fall, the still-warm Great Lakes give heat and moisture to cold air passing over them. That is part of what causes lake-effect snow squalls east of Lake Huron and Georgian Bay.
- Our bodies are mostly water, which helps keep our temperature steady.
- Water is used as a coolant in car engines and power plants because it can carry away a lot of energy for each degree it warms.
Just for funWater is the friend at the party who takes forever to warm up to people, but once warmed up, stays warm all night. Its specific heat capacity is about 4.6 times that of the aluminium foil wrapped around your leftovers (4180 ÷ 900 ≈ 4.6).
Hands-on activityWater versus oil: which changes temperature faster?
You need: a kitchen scale, two identical cups (same material and size), tap water, vegetable oil, a kitchen thermometer and a fridge.
- Put each empty cup on the scale and zero it. Add 150 g of water to one cup and 150 g of oil to the other. Leave both on the counter for 30 minutes so they start at room temperature. Record each starting temperature.
- Put both cups in the fridge side by side for 20 minutes.
- Take them out and immediately measure each temperature (stir gently with the thermometer). Wipe the thermometer between cups.
- Calculate ΔT for each. Which liquid changed more? Which has the larger specific heat capacity?
- Using c(water) = 4180 J/(kg·°C), calculate how much heat the water lost. Assuming the oil lost about the same amount of heat (same fridge, same cup, same time), estimate c for the oil with c = Q/(mΔT).
What to look for: the oil's temperature drops more, so it has the smaller c. Published values for cooking oils are roughly 2000 J/(kg·°C); your estimate may differ because the two liquids do not lose exactly the same heat, the cups also cool, and the thermometer takes time to respond. Name one of these as a source of error in your conclusion.
Your turn
1How much heat is needed to raise the temperature of 3.0 kg of water by 25 °C?
Q = mcΔT = (3.0)(4180)(25) = 313 500 J ≈ 3.1 × 10⁵ J.
2A 2.0 kg iron skillet warms from 20 °C to 220 °C. How much heat does it absorb? (c = 450 J/(kg·°C))
ΔT = 200 °C. Q = (2.0)(450)(200) = 1.8 × 10⁵ J (180 kJ).
3250 g of hot chocolate (treat it as water) cools from 70 °C to 40 °C. Find Q.
m = 0.250 kg, ΔT = −30 °C. Q = (0.250)(4180)(−30) = −31 350 J ≈ −3.1 × 10⁴ J (it loses about 31 kJ).
4A 0.40 kg sample absorbs 3600 J and warms by 10 °C. Find c and identify the substance.
c = Q/(mΔT) = 3600 ÷ (0.40 × 10) = 900 J/(kg·°C): aluminium.
5What mass of water can be heated from 20 °C to 80 °C by 5.0 × 10⁵ J?
m = Q/(cΔT) = 500 000 ÷ (4180 × 60) = 500 000 ÷ 250 800 ≈ 2.0 kg.
6A 1.2 kg copper pot at 20 °C absorbs 4.68 × 10⁴ J. What is its final temperature?
ΔT = Q/(mc) = 46 800 ÷ (1.2 × 390) = 46 800 ÷ 468 = 100 °C. T₂ = 20 + 100 = 120 °C.
7What is the heat capacity of 5.0 kg of water?
Heat capacity = mc = 5.0 × 4180 = 2.1 × 10⁴ J/°C (20 900 J for every 1 °C).
More practice: warm-up
1Write the units of specific heat capacity.
3A sample cools from 60 °C to 25 °C. What is ΔT?
4If Q is negative, did the substance gain or lose heat?
5Which needs more energy to warm by 1 °C: 1 kg of water or 1 kg of iron?
1 kg of water (4180 J versus 450 J).
More practice: on your own
1A hot-water tank holds 180 kg of water. How much energy is needed to heat it from 10 °C to 60 °C?
Q = (180)(4180)(50) = 37 620 000 J ≈ 3.8 × 10⁷ J (38 MJ).
2A 50.0 g lead fishing sinker absorbs 325 J. By how much does its temperature rise? (c = 130 J/(kg·°C))
ΔT = Q/(mc) = 325 ÷ (0.0500 × 130) = 325 ÷ 6.5 = 50 °C.
3Explain why a 1 kg block of aluminium and a 1 kg block of copper, both taken from boiling water, would not hold the same amount of thermal energy relative to room temperature.
They have the same mass and the same temperature change to room temperature, but aluminium's c (900) is more than twice copper's (390). By Q = mcΔT, the aluminium gives off more than twice as much heat as it cools.
4A 0.300 kg glass dish cools from 175 °C to 25 °C. How much heat does it lose? (c = 840 J/(kg·°C))
Q = (0.300)(840)(−150) = −3.78 × 10⁴ J. It loses about 37.8 kJ.
5Why are cities beside the Great Lakes often slower to warm up in spring than cities farther inland?
The lakes are huge masses of water with a high specific heat capacity, so they warm slowly. Cool air over the cold lakes keeps nearby land cooler in spring.
Challenge
1An electric kettle supplies 1.5 × 10³ J every second to 1.2 kg of water at 18 °C. Ignoring losses, how long does it take to reach 100 °C?
Q = (1.2)(4180)(82) = 411 312 J. Time = 411 312 ÷ 1500 ≈ 274 s, about 4.6 minutes.
2Equal masses of water and aluminium receive equal amounts of heat. The water warms by 6.0 °C. By how much does the aluminium warm?
Same Q and m, so ΔT is inversely proportional to c: ΔT(Al) = 6.0 × (4180 ÷ 900) ≈ 28 °C.
3A 0.80 kg aluminium pot contains 2.0 kg of water, both at 15 °C. How much heat is needed to bring the pot and water to 100 °C? What fraction of this energy goes into the pot?
Water: (2.0)(4180)(85) = 710 600 J. Pot: (0.80)(900)(85) = 61 200 J. Total = 771 800 J ≈ 7.7 × 10⁵ J. Fraction for the pot = 61 200 ÷ 771 800 ≈ 0.079, about 8%.
Quiz
1. The specific heat capacity of water is 4.18 × 10³ J/(kg·°C). What does this mean?
2. How much heat is needed to warm 2.0 kg of water by 10 °C?
3. Equal masses of copper (c = 390) and water (c = 4180) receive the same heat. Which is true?
4. A substance cools from 90 °C to 30 °C. Using the sign convention in this unit, Q is:
5. A 0.50 kg block absorbs 4500 J and warms by 20 °C. What is its specific heat capacity?
6. Why do coastal and Great Lakes communities often have milder winters than inland communities?
Big idea: Q = mcΔT (expectation D2.10) is the first heat equation, and specific heat capacity is a key concept in D3.3. Students should be comfortable rearranging it and using a sign convention.
How to help: Ask your teen to estimate how much energy it takes to heat a full kettle or a hot-water tank, then check with the equation. Have them explain why the bathwater stays warm a long time, or why the sand cools faster than the lake at night.
Common difficulties:
- Forgetting to convert grams to kilograms (answers come out 1000 times too big).
- Calculating ΔT as initial − final, which flips the sign.
- Confusing specific heat capacity (per kilogram, a property of the substance) with heat capacity (for a whole object).
- Using Q = mcΔT across a change of state. It only works while the substance stays solid, liquid or gas.
- Rounding too early, or keeping too many digits. Round only the final answer.