Specific heat capacity: ΔE = mcΔθ
This unit is A Level (IA2) content, and it assumes the AS work on energy, power, efficiency and density from Unit 1 and on electrical power from Unit 2.
It is examined in the Unit 5 paper (WPH15), where thermodynamics has carried about a fifth of the marks in recent series, nearly always as a real-world heating problem with a power, a time and a claim to test.
Its three core practicals (CP12, CP13 and CP14) are also examined in the Unit 6 practical paper (WPH16).
This first section covers the first of two equations, the equation ; the next section covers .
- Each marking point is worth one mark. Our answers write it as "(1)"; the physics mark schemes use no M1, A1 or B1 codes.
- A calculation earns a "use of" mark only when the equation appears with the values substituted; a rearranged symbol equation on its own earns nothing. Then a mark for the answer with its unit. A missing or wrong unit loses that last mark (not on a show that, and not where the unit is printed).
- Error carried forward (ecf): a wrong value from an earlier part is followed through, so always carry on.
- Show that: give your final value to at least one more significant figure than the printed value.
- Deduce, Assess, Evaluate: the last mark is for comparing two values and stating a conclusion in words. Criticise: a mark for saying exactly where the statement is wrong.
- The one asterisked 6-mark question counts indicative points (up to 4 marks) plus up to 2 marks for a linked, logical line of reasoning.
- Temperature in kelvin: . A temperature difference is the same number in K and in °C.
- Volume and surface area of a sphere: and .
- Moles: , with mol⁻¹, and the molar gas constant J mol⁻¹ K⁻¹, if you use . Neither nor is on the data sheet.
- The potential-divider ratio (for CP12): .
- Time conversions: 1 h = 3600 s, 1 day = 86 400 s.
On the list (Unit 5 block): , , , .
The Boltzmann constant J K⁻¹ is on the data sheet.
- Unit 1: density (mass of water from a volume), power and efficiency , all on the list.
- Unit 2: electrical power and energy , on the list; potential dividers and thermistors for core practical 12.
The idea: same energy, different temperature rise
Put the same electric heater in 1 kg of water and then in 1 kg of cooking oil for one minute each.
The oil's temperature rises about twice as much, though it received exactly the same energy.
Different substances need different amounts of energy to warm by the same amount.
The quantity that captures this is the specific heat capacity, .
Why should substances differ?
Energy supplied by heating goes into the random motion of the particles (their kinetic energy) and into stretching the bonds between them (their potential energy); section 5 calls this the internal energy.
A kilogram of water contains far more molecules than a kilogram of copper atoms, and each water molecule can also rotate and vibrate.
So the same energy, shared among more particles and more ways of moving, raises the average kinetic energy, and so the temperature, less.
The energy needed is proportional to three things:
- the mass : twice the water needs twice the energy for the same rise;
- the temperature change : a 20 K rise needs twice the energy of a 10 K rise;
- the material, through .
Put together:
Here is the energy transferred in J, the mass in kg, the temperature change in K or °C, and the specific heat capacity in J kg⁻¹ K⁻¹.
The equation is on the list.
It applies only while the substance stays in one state; at a melting or boiling point the temperature stops changing and the next section's equation takes over.
- Specific heat capacity: the energy required to raise the temperature of 1 kg of a substance by 1 K.
- Temperature change : the final temperature minus the initial temperature; it has the same value in kelvin as in degrees Celsius.
Reading as a gradient
If a block is heated at a steady rate and nothing is lost, its temperature rises in a straight line.
On a graph of temperature against energy supplied, the gradient is .
The graph below compares equal masses of aluminium ( J kg⁻¹ K⁻¹) and water ( J kg⁻¹ K⁻¹).
After 40 kJ the aluminium is about 70 K hotter than the water.
If instead you plot temperature against time for a heater of power , the energy supplied is , so the gradient is and .
A real block loses some energy to the air, so its line rises a little less steeply than this, the measured gradient is too small, and the value of calculated from it comes out too large.
Some typical values (to 2 or 3 s.f.):
Substance
/ J kg⁻¹ K⁻¹
water
4180
ice
2100
cooking oil
about 2000
aluminium
900
concrete
about 880
copper
385
lead
128
Water's very large value is why it is used in central heating and car cooling systems: each kilogram carries a lot of energy for a small temperature change.
It is also why coastal climates are milder: the sea warms and cools far less than the land in a day.
For many simple solids, multiplied by the mass of one mole is roughly the same, about 25 J mol⁻¹ K⁻¹.
So, per particle, solids need about the same energy per kelvin; per kilogram is small for lead only because a kilogram of lead contains few, heavy atoms.
This links to the molecular picture of the later sections.
Connections
Specific heat capacity is the start of a chain the papers use again and again: an electrical power from Unit 2 (), an efficiency from Unit 1, an energy from and a time from .
Engineers size heaters, cooling systems and storage tanks this way; the same balance of energy in against energy stored decides how fast a spacecraft's instruments warm in sunlight, an application named for this topic.
Method and exam technique
Every calculation in this unit is set out the way a mark scheme credits it: the equation chosen, the substitution shown with every value in SI units, then the answer with a unit, each on its own line.
An aluminium saucepan of mass 1.5 kg is heated from 18 °C to 95 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹.
Calculate the energy transferred to the saucepan.
(2)
1. K (a difference, so no 273 is added)
2. (1) use of
3. J
Answer: J (1) answer with unit.
About 100 kJ: the same as lifting a 100 kg mass by about 100 m, which is why cooking uses so much energy.
An electric kettle has a power of 2.2 kW.
It contains 0.75 kg of water at 14 °C, and 85% of the electrical energy supplied is transferred to the water.
The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹.
Calculate the time taken for the water to reach 100 °C. (4)
1. J (1) use of
2. Energy supplied J (1) efficiency used the right way round: the useful energy is the 85%
3. (1) use of
4. s
Answer: 140 s (1) answer with unit, about two and a half minutes.
Dividing by 0.85 makes the time longer, which is a quick sense check: losses always mean more energy and more time.
A student heats a 0.800 kg copper block with an electric heater connected to a 12.0 V supply.
The current is 4.00 A, and in 300 s the temperature of the block rises from 19.2 °C to 66.0 °C. Show that the specific heat capacity of copper is about 400 J kg⁻¹ K⁻¹.
(3)
1. J (1) use of (or with )
2. (1) use of
3. J kg⁻¹ K⁻¹ (1) value to at least 2 s.f., one more than the printed "400"
Answer: 385 J kg⁻¹ K⁻¹, about 400.
Notice the student's value agrees with copper's accepted value only because this example ignores losses.
In a real lesson some energy heats the air, the block reaches a lower temperature, and the calculated comes out higher than 385.
A garden hot tub holds 1.5 m³ of water at 15 °C. Its 3.0 kW heater runs continuously, but only 80% of the energy it supplies stays in the water; the rest is lost through the cover and walls.
The seller claims the water will be ready at 38 °C "overnight, in 12 hours".
The density of water is 1000 kg m⁻³ and its specific heat capacity is 4180 J kg⁻¹ K⁻¹.
Deduce whether the claim is correct.
(5)
1. kg (1) use of (synoptic, Unit 1)
2. J (1) use of
3. Useful power W (1) the fraction that stays in the water
4. s h (1) use of , converted to hours
5. 16.7 h is longer than the 12 h claimed, so the claim is not correct (1) comparison of two values and a conclusion
Answer: the tub takes about 17 hours, so it is not ready overnight; the claim is wrong.
The conclusion line quotes both times.
"So it takes 16.7 hours" alone would lose the last mark, the one the examiner reports say is dropped most often.
In a heating chain, write a short label beside each line ("energy to warm the water", "useful power", "time") so the examiner can follow it, and keep energy (J) and power (W) apart: an energy divided by a power is a time, never the other way round.
If the question gives a percentage efficiency or loss, decide in words which part is useful before you use it.
- Adding 273 to a temperature change. A rise from 15 °C to 38 °C is 23 K, not 296 K. Converting to kelvin is for an absolute temperature in , never for . Students do it because "always use kelvin" sounds like a safe rule; in the 273s cancel.
- Using the efficiency the wrong way round. If 80% of the energy stays in the water, the useful power is ; multiplying the energy needed by 0.80 instead makes the time shorter, which cannot be right.
- Mixing energy and power. Writing "power " or dividing a power by an energy. Check the unit of every line: J, W or s.
- Volume in the wrong unit. A volume in litres or cm³ must become m³ before : 1 L m³, 1 cm³ m³.
- Combining every step in one calculator line. One slip then loses every process mark; show each substitution.
Check your understanding
1. Blocks X and Y have equal masses. Each is supplied with the same energy, and the temperature rise of X is three times the temperature rise of Y. Which of the following is correct?
A
B
C
D
(1) Total [1]
2. A cup holds 0.35 kg of milk at 5 °C. It is heated to 65 °C. The specific heat capacity of milk is 3900 J kg⁻¹ K⁻¹. Calculate the energy needed. (2) Total [2]
3. A 750 W heater warms 2.0 kg of cooking oil by 40 K. Only 90% of the energy supplied is transferred to the oil. The specific heat capacity of the oil is 1970 J kg⁻¹ K⁻¹. Show that the time taken is about 4 minutes. (3) Total [3]
4. A student writes: "The temperature of the water rose from 20 °C to 30 °C, so in kelvin the rise is from 293 K to 303 K, which is 303 K." Explain what is wrong with this and give the correct temperature change. (2) Total [2]
5. Explain why the temperature of the sea changes much less than the temperature of a sandy beach during a sunny day, though both receive about the same energy per square metre. (2) Total [2]
6. A liquid of mass 0.50 kg is heated by a 60 W heater. The graph shows the temperature of the liquid against time.
(a) Determine the gradient of the graph. (2)
(b) Calculate the specific heat capacity of the liquid. (2)
(c) The accepted value for this liquid is 2400 J kg⁻¹ K⁻¹. Suggest why the student's value is larger. (2)
Total [6]
Answers
1. B (1). For the same and , , so a three times larger means a one third as large. A inverts the relationship; C and D square the factor, as if depended on .
2. (1); J (1) (81 900 J).
3. J (1); useful power W and (1); s minutes (1) (at least 2 s.f.; 3.9 min or 233 s, both one more figure than "about 4").
4. A temperature change is a difference, and adding 273 to both temperatures cancels in the subtraction (1); the change is K, the same as 10 °C, not 303 K (1).
5. Water has a much larger specific heat capacity than sand (1); so the same energy per kilogram gives a much smaller temperature rise, since (1). (Also creditable: the sea's energy is spread through a much larger mass of water by mixing.)
6. (a) Large triangle from to (1); gradient K s⁻¹ (1) (accept 0.044 to 0.046). (b) (1) use of with ; J kg⁻¹ K⁻¹ (1) (2670; ecf from (a)). (c) Some of the heater's energy is transferred to the surroundings (and the container) (1); so the temperature rises less than it should, the gradient is too small and comes out too large (1).