Answers · Specific heat capacity

Measuring the specific heat capacity of water — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. HQ11 · Practice

Model answer. (a) Each cubic centimetre of water has a mass of 1 g, so 250 cm³ has a mass of 250 g. (b) There are 1000 g in 1 kg, so 250 g = 250 ÷ 1000 = 0.250 kg.

Checked automatically

(a) Mass in grams: 250 g (exact value)
- Common wrong answer 0.25 : That is the mass in kilograms, which (b) asks for. In grams: 1 g for each cm³, so 250 g.
(b) Mass in kilograms: 0.250 kg (exact value)
- Common wrong answer 250000 : To change grams into kilograms, divide by 1000: 250 ÷ 1000 = 0.250 kg.
- Common wrong answer 2.5 : There are 1000 g in 1 kg, not 100: 250 ÷ 1000 = 0.250 kg.
- Common wrong answer 250 kg: Each cubic centimetre of water is 1 g, not 1 kg. 250 g ÷ 1000 = 0.250 kg.

A full answer includes

(a) 250 g.
(b) 250 g ÷ 1000 = 0.250 kg.

Also accept

(a) 250 g; (b) 0.250 kg.

Watch for

“(b) 250 cm³ of water has a mass of 250 kg.” — Each cubic centimetre has a mass of 1 g, not 1 kg: 250 cm³ is 250 g, which is 0.250 kg.
“(b) 250 g = 2.5 kg.” — There are 1000 g in a kilogram, not 100: divide by 1000.

2. HQ12 · Practice

Model answer. So that all the water is at the same temperature as the thermometer. The water next to the heater warms first. Without stirring, some parts of the water are hotter than others, and the thermometer shows only the temperature around its bulb, which could be higher or lower than the rest. The calculation needs the rise in temperature of all the water, so the water is stirred to keep it at one temperature.

A full answer includes

I chose “So that all the water is at the same temperature as the thermometer”.
I explained that without stirring the water near the heater is hotter than the rest, so the thermometer shows only the temperature around its bulb, not the rise in temperature of all the water.

Also accept

so that all the water is at the same temperature as the thermometer; stirring keeps all the water at one temperature, so the reading is the temperature of all of it.

Watch for

“Stirring makes the water heat up faster.” — The heater supplies the same energy whether you stir or not. Stirring only spreads it evenly through the water.
“Stirring adds energy to the water.” — Gentle stirring adds so little energy that it does not matter. Its job is to mix the water so that it all has one temperature.

3. HQ13 · Practice

Options as numbered on the pupil sheet: 1 — Set the joulemeter to zero, switch on and stir the water gently; 2 — Measure out 1000 cm³ of water (1.00 kg) and pour it into the lagged beaker; 3 — Switch off when the joulemeter shows 30.0 kJ, stir, and read the highest temperature; 4 — With the power off, put the heater and the thermometer into the water, terminals above it, and read the starting temperature.

Model answer. First measure out 1000 cm³ of water, which has a mass of 1.00 kg, into the lagged beaker. With the power off, put the heater and the thermometer into the water, with the heater’s terminals above the surface, and read the starting temperature. Set the joulemeter to zero, switch on and keep stirring gently. When the joulemeter shows 30.0 kJ, switch off, go on stirring and read the highest temperature. Then Δθ is the highest temperature minus the starting temperature, and c = 30 000 J ÷ (1.00 kg × Δθ).

A full answer includes

Step 1: measure out 1000 cm³ of water (1.00 kg) into the lagged beaker.
Step 2: with the power off, put the heater and the thermometer in, terminals above the water, and read the starting temperature.
Step 3: set the joulemeter to zero, switch on and stir.
Step 4: switch off at 30.0 kJ, stir, and read the highest temperature; then work out Δθ and c.

Also accept

measure out 1000 cm³ of water into the lagged beaker; power off, heater and thermometer in, read the starting temperature; joulemeter to zero, switch on and stir; switch off at 30.0 kJ, stir, read the highest temperature, work out Δθ and c.

Watch for

“Switch on first, then lower the heater into the water.” — Never switch on a heater that is out of the water: in air it gets very hot very quickly. Put it in with the power off, then switch on.
“Read the final temperature the moment you switch off, without stirring.” — Keep stirring and read the highest temperature: the water near the heater is still sharing its energy with the rest.

4. HQ14 · Practice

Model answer. ΔE = 8.4 kJ = 8400 J. m = 200 g = 0.200 kg. Δθ = 30.0 °C − 20.0 °C = 10.0 °C. c = ΔE / (mΔθ) = 8400 J ÷ (0.200 kg × 10.0 °C) = 8400 ÷ 2.00 = 4200 J/(kg °C).

Checked automatically

Specific heat capacity of water: 4200 J/(kg °C) (within ±2%)
- Common wrong answer 4.2 J/(kg °C): A thousand times too small: you used 8.4 instead of 8400 J, or 200 instead of 0.200 kg. Change kJ into J and g into kg first.
- Common wrong answer 0.0042 : You made both slips: change 8.4 kJ into 8400 J and 200 g into 0.200 kg.
- Common wrong answer 1400 : You divided by the final temperature, 30.0 °C. Δθ is the rise: 10.0 °C.
- Common wrong answer 16800 : You multiplied. c = ΔE ÷ (m × Δθ): divide the energy by the mass and by the rise.

A full answer includes

I used ΔE = 8400 J, m = 0.200 kg and Δθ = 10.0 °C.
I calculated c = 8400 J ÷ (0.200 kg × 10.0 °C) = 4200 J/(kg °C), with the unit.

Also accept

4200 J/(kg °C).

Watch for

“c = 8400 ÷ (200 × 10.0) = 4.2 J/(kg °C).” — The mass must be in kilograms: 200 g = 0.200 kg.
“c = 8400 ÷ (0.200 × 30.0) = 1400 J/(kg °C).” — 30.0 °C is the final temperature. The rise is 30.0 °C − 20.0 °C = 10.0 °C.

5. HQ15 · Practice

Options as numbered on the pupil sheet: 1 — water (4200 J/(kg °C)); 2 — copper (390 J/(kg °C)); 3 — dense bricks (about 800 J/(kg °C)); 4 — polystyrene (a good insulator); 5 — dry sand (about 830 J/(kg °C)).

Model answer. Coolant: water. Each kilogram takes in about 4200 J for every degree it warms, so it carries a lot of energy away from the engine while warming by only a few degrees. Saucepan base: copper. With only 390 J/(kg °C), little of the energy from the hob is used up warming the pan itself. Storage heater core: dense bricks. A large mass of brick can be heated to several hundred degrees Celsius and stores a lot of energy (ΔE = mcΔθ with a large m and a large Δθ); water would boil at 100 °C. Polystyrene is a good insulator, which is why things are wrapped in it, but it would melt in a storage heater.

A full answer includes

Coolant: water, because its high specific heat capacity lets each kilogram carry away a lot of energy for a small rise in temperature.
Saucepan base: copper, because its low specific heat capacity means little energy is used up warming the pan itself.
Storage heater core: dense bricks, because a large mass of them can be heated to several hundred °C, far above the 100 °C at which water boils.

Also accept

coolant – water; saucepan base – copper; storage heater core – dense bricks.

Watch for

“A saucepan should have a high specific heat capacity, so that it stores lots of energy.” — A pan is there to pass energy on to the food. With a low specific heat capacity, less energy is used up warming the pan itself, so the food heats sooner.
“The storage heater core should be water, because water has the highest specific heat capacity.” — Water boils at 100 °C, but storage heater bricks are heated to several hundred degrees. A large mass and a large rise in temperature store more energy than water could, even though water’s c is higher.

6. HQ16 · Practice

Model answer. 1.50 litres = 1500 cm³, so the water has a mass of 1500 g = 1.50 kg. Δθ = 100 °C − 20 °C = 80 °C. ΔE = mcΔθ = 1.50 kg × 4200 J/(kg °C) × 80 °C = 504 000 J = 504 kJ. A real kettle needs more, because some energy warms the kettle and some escapes to the air.

Checked automatically

Energy needed: 504 kJ (within ±2%)
- Common wrong answer 630 : You used the final temperature, 100 °C. The rise is 100 − 20 = 80 °C.
- Common wrong answer 504000 kJ: 504 000 is the energy in joules, or your mass was in grams. With the mass as 1.50 kg, 504 000 J ÷ 1000 = 504 kJ.

A full answer includes

m = 1500 cm³ of water = 1500 g = 1.50 kg; Δθ = 100 °C − 20 °C = 80 °C.
ΔE = 1.50 kg × 4200 J/(kg °C) × 80 °C = 504 000 J = 504 kJ.

Also accept

504 kJ.

Watch for

“ΔE = 1.50 × 4200 × 100 = 630 kJ.” — 100 °C is the final temperature; the water started at 20 °C. Use the rise: 80 °C.
“m = 1500 kg.” — 1500 cm³ of water is 1500 g, which is 1.50 kg.

7. HQ17 · Practice

Model answer. Sand has a lower specific heat capacity, so the same energy gives each kilogram a bigger rise in temperature. Each kilogram of dry sand needs about 830 J for each degree, each kilogram of water about 4200 J: 4200 ÷ 830 ≈ 5. So the same energy warms a kilogram of sand about five times as much as a kilogram of water. The sea also mixes, sharing its energy with the deeper water.

A full answer includes

I chose “Sand has a lower specific heat capacity, so the same energy gives each kilogram a bigger rise in temperature”.
I compared the values: water needs about 4200 ÷ 830 ≈ 5 times as much energy per kilogram for each degree, so the same energy warms sand about five times as much.

Also accept

sand has a lower specific heat capacity; water needs about five times as much energy per kg per °C (4200 against 830), so the sand warms about five times as much.

Watch for

“Sand has a higher specific heat capacity, so it takes in more energy and gets hotter.” — A high specific heat capacity means a lot of energy is needed for each degree, so the temperature rises slowly. Sand’s is the lower one, so it warms quickly.
“The sea is heavier, so it needs more energy.” — Specific heat capacity compares equal masses: even kilogram for kilogram, water needs about five times as much energy for each degree as sand.

8. HQ18 · Practice

Model answer. ΔE = 21 kJ = 21 000 J. (a) Water: Δθ = ΔE / (mc) = 21 000 J ÷ (1.00 kg × 4200 J/(kg °C)) = 5.0 °C. (b) Copper: Δθ = 21 000 J ÷ (1.00 kg × 390 J/(kg °C)) ≈ 53.8 °C, about 54 °C. The copper warms more than ten times as much, because each kilogram of it needs far less energy for each degree.

Checked automatically

(a) Rise in temperature of the water: 5.0 °C (within ±2%)
- Common wrong answer 0.005 : The energy must be in joules: 21 kJ = 21 000 J.
- Common wrong answer 88200000 : You multiplied. Δθ = ΔE ÷ (m × c): divide the energy by 1.00 kg × 4200 J/(kg °C).
(b) Rise in temperature of the copper: 53.8 °C (within ±2%)
- Common wrong answer 0.0538 : The energy must be in joules: 21 kJ = 21 000 J.
- Common wrong answer 8190000 : You multiplied. Δθ = ΔE ÷ (m × c): divide the energy by 1.00 kg × 390 J/(kg °C).

A full answer includes

(a) Δθ = 21 000 J ÷ (1.00 kg × 4200 J/(kg °C)) = 5.0 °C.
(b) Δθ = 21 000 J ÷ (1.00 kg × 390 J/(kg °C)) ≈ 54 °C: more than ten times the rise of the water.

Also accept

(a) 5.0 °C; (b) about 54 °C.

Watch for

“(a) Δθ = 21 ÷ 4200 = 0.005 °C.” — The energy must be in joules: 21 kJ = 21 000 J.
“The same energy gives the same rise in temperature in both.” — Each material needs its own energy for each degree. Copper needs far less than water, so its temperature rises far more.

9. HQ19 · Practice

Model answer. At 21 kJ, Q has risen from 20 °C to 30 °C, a rise of 10 °C: c = 21 000 J ÷ (0.500 kg × 10 °C) = 4200 J/(kg °C), so Q is water. Its line is the less steep one, because water needs more energy for each degree. P has risen from 20 °C to 40 °C, a rise of 20 °C: c = 21 000 J ÷ (0.500 kg × 20 °C) = 2100 J/(kg °C), half that of water. The same energy gave P twice the rise.

Checked automatically

Specific heat capacity of the other liquid: 2100 J/(kg °C) (within ±3%)
- Common wrong answer 1050 : You used P’s final temperature, 40 °C, or left out the mass. Use the rise, 20 °C, and divide by m × Δθ = 0.500 kg × 20 °C.
- Common wrong answer 4200 : 4200 J/(kg °C) is water, liquid Q. The question asks for the other liquid, P.
- Common wrong answer 2.1 J/(kg °C): A thousand times too small: change 21 kJ into 21 000 J first.
- Common wrong answer 210000 : You multiplied. c = ΔE ÷ (m × Δθ).

A full answer includes

Q is water: 21 000 J ÷ (0.500 kg × 10 °C) = 4200 J/(kg °C); its line is the less steep one.
P: Δθ = 40 °C − 20 °C = 20 °C, so c = 21 000 J ÷ (0.500 kg × 20 °C) = 2100 J/(kg °C).

Also accept

2100 J/(kg °C); Q is water, the less steep line.

Watch for

“P is water, because its temperature rises more.” — Water has a high specific heat capacity, so the same energy gives it a smaller rise. The less steep line, Q, is water.
“c of P = 21 000 ÷ (0.500 × 40) = 1050 J/(kg °C).” — 40 °C is P’s final temperature. Use its rise from 20 °C: 20 °C.

10. HQ20 · Practice

Model answer. The joulemeter counted all the energy supplied, but not all of it warmed the water. With no lagging and no lid, a lot escaped to the air through the sides of the beaker and from the open surface, and some warmed the beaker, the heater and the thermometer. So the water’s temperature rose less than it would have done, and a smaller Δθ in c = ΔE / (mΔθ) gives a bigger c. Improvements: wrap the beaker in lagging and put a lid on it, or use a polystyrene cup with a lid instead of a glass beaker; and stir all the time, so that the thermometer reads the temperature of all the water.

A full answer includes

I explained that energy escaped to the air (through the sides and from the open surface) and warmed the beaker, the heater and the thermometer, so the water’s rise in temperature was smaller than it should have been.
I said that a smaller Δθ in c = ΔE / (mΔθ) gives a bigger c.
I suggested two improvements, such as wrapping the beaker in lagging, putting on a lid, or using a polystyrene cup with a lid (and stirring all the time).

Also accept

energy escaped to the air and warmed the beaker, so Δθ was too small; lag the beaker and add a lid.

Watch for

“Heat the water for longer to get a better value.” — A longer run makes the water hotter, so it loses energy to the air even faster. Insulating the beaker is what keeps the energy in.
“The value is high because water has a very high specific heat capacity.” — Water’s real value is 4200 J/(kg °C). The extra comes from the method: energy that never reached the water was still counted by the joulemeter.

11. HT2 · Apply

Model answer. (a) Water has a very high specific heat capacity, about 4200 J/(kg °C): each kilogram carries away about 4200 J for every degree it warms, so a modest flow of water takes a lot of energy from the engine while warming by only a few degrees. (b) Δθ = 90 °C − 80 °C = 10 °C. ΔE = mcΔθ = 5.0 kg × 4200 J/(kg °C) × 10 °C = 210 000 J = 210 kJ. (c) Nearer 20 °C. The energy given out by the aluminium as it cools is taken in by the water as it warms. Each kilogram of water needs about 4.7 times as much energy for each degree as aluminium, so the water warms only a little while the aluminium cools a lot, until both are at the same temperature (in fact about 31 °C). (d) Weigh an empty beaker, add the liquid and weigh again to find its mass in kg. Wrap the beaker in lagging, put the heater and the thermometer in with the power off, and read the starting temperature. Set the joulemeter to zero, switch on and stir all the time. Switch off after a known energy, stir and read the highest temperature. Then c = ΔE / (mΔθ). Safety: use a low-voltage supply, never the mains; switch off before moving the heater; keep the heater’s terminals above the liquid.

A full answer includes

(a) Its high specific heat capacity: each kilogram carries away a lot of energy (about 4200 J for each degree) for a small rise in temperature.
(b) ΔE = 5.0 kg × 4200 J/(kg °C) × 10 °C = 210 000 J = 210 kJ.
(c) Nearer 20 °C: the energy the aluminium gives out is taken in by the water, and each kilogram of water needs about 4.7 times as much energy for each degree, so the aluminium cools far more than the water warms.
(d) Find the mass of the liquid (weigh the beaker empty and full); in a lagged beaker put the heater and the thermometer in and read the start; zero the joulemeter, heat and stir; read ΔE and the final temperature; c = ΔE / (mΔθ). Precaution: a low-voltage supply, power off before moving the heater, terminals above the liquid.

Also accept

(a) high specific heat capacity, so lots of energy carried for a small rise; (b) 210 kJ; (c) nearer 20 °C, water needs about 4.7 times the energy per °C; (d) mass, energy and rise in a lagged, stirred beaker, c = ΔE/(mΔθ); low voltage, power off before moving the heater.

Watch for

“(c) The final temperature is 50 °C, halfway, because the masses are equal.” — Equal masses are not enough: the water needs about 4.7 times as much energy for each degree. The energy the aluminium gives out warms the water only a little, so the final temperature is much nearer 20 °C.
“(a) Water is a good coolant because it is cold.” — The water in an engine is hot, around 80 to 90 °C. It works because each kilogram carries a lot of energy for each degree it warms: its specific heat capacity is high.