Answers · Specific heat capacity

Specific heat capacity — return questions key

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

1. HR1 · Return

Model answer. (a) The internal energy of the object increases. (b) The particles stay in their fixed positions but vibrate faster about them, so the average kinetic energy of all the particles increases. (c) The specific heat capacity of a material is the energy required per unit mass per unit temperature increase: the energy needed to raise the temperature of 1 kg of it by 1 °C, c = ΔE / (mΔθ). (d) Δθ = 45 °C − 20 °C = 25 °C. ΔE = mcΔθ = 0.40 kg × 900 J/(kg °C) × 25 °C = 9000 J. (e) The joulemeter counts all the energy supplied, but some of it escapes from the block to the air and some warms the heater and the thermometer. The block’s temperature rises less than it would if all the energy stayed in it, and a smaller Δθ in c = ΔE / (mΔθ) gives a bigger c. (f) Weigh the block on a balance to find m. Put an immersion heater in one hole and a thermometer in the other, wrap the block in insulation, set the joulemeter to zero and read the starting temperature. Switch on, heat until the joulemeter shows a chosen energy, switch off and read the highest temperature. Then Δθ = final − starting temperature, ΔE is the joulemeter reading in joules, and c = ΔE / (mΔθ).

A full answer includes

(a) It increases.
(b) They vibrate faster about their fixed positions: their average kinetic energy increases.
(c) The energy required per unit mass per unit temperature increase (the energy to warm 1 kg by 1 °C).
(d) ΔE = 0.40 kg × 900 J/(kg °C) × 25 °C = 9000 J.
(e) Some energy escapes to the air (and some warms the heater and the thermometer), so the rise is smaller than it should be, which gives a bigger c.
(f) Mass on a balance; starting and final temperature with a thermometer in the block; energy supplied with a joulemeter set to zero first (from an immersion heater in the block); the block insulated; c = ΔE / (mΔθ) with Δθ = final − starting temperature.

Also accept

(a) increases; (b) vibrate faster, higher average kinetic energy; (c) energy per unit mass per unit temperature increase; (d) 9000 J; (e) energy escapes, so Δθ is too small; (f) m on a balance, θ before and after with a thermometer, ΔE from a joulemeter, insulation, c = ΔE / (mΔθ).

Watch for

“(d) ΔE = 0.40 × 900 × 45 = 16 200 J.” — 45 °C is the final temperature. Use the rise: 45 °C − 20 °C = 25 °C.
“(c) Specific heat capacity is the energy needed to heat something up.” — That leaves out what makes the comparison fair: the energy is per kilogram and per degree. c is the energy required per unit mass per unit temperature increase.

2. HR2 · Return

Model answer. (a) 1 cm³ of water has a mass of 1 g, so 1000 cm³ has a mass of 1000 g = 1.00 kg. (b) Measure out 1000 cm³ of water (1.00 kg) into a lagged beaker. With the power off, put the immersion heater and the thermometer in, and read the starting temperature. Set the joulemeter to zero, switch on and stir all the time. After a known energy, such as 30.0 kJ, switch off, stir and read the highest temperature. Work out Δθ, change the energy into joules and calculate c = ΔE / (mΔθ). (c) Stirring keeps all the water at one temperature, the temperature the thermometer shows. Lagging cuts the energy escaping to the air, so that more of the counted energy stays in the water and the rise is not too small. (d) Δθ = 40 °C − 20 °C = 20 °C. ΔE = mcΔθ = 0.50 kg × 4200 J/(kg °C) × 20 °C = 42 000 J. (e) The energy stored is ΔE = mcΔθ. Bricks have a lower specific heat capacity than water, but a large mass of them can be heated to several hundred degrees Celsius, far above the 100 °C at which water would boil, so a large m and a large Δθ store a lot of energy. The insulation round them lets it out slowly during the day.

A full answer includes

(a) 1.00 kg (1 cm³ of water has a mass of 1 g).
(b) Measure the mass of the water; put the heater and the thermometer in and read the starting temperature; zero the joulemeter and heat, stirring; read the energy and the final temperature; c = ΔE / (mΔθ).
(c) Stirring keeps all the water at the temperature the thermometer shows; lagging stops energy escaping to the air, so the rise is not too small.
(d) ΔE = 0.50 kg × 4200 J/(kg °C) × 20 °C = 42 000 J.
(e) A large mass and a large rise in temperature store a lot of energy (ΔE = mcΔθ); bricks can be heated far above 100 °C, where water would boil.

Also accept

(a) 1.00 kg; (b) mass, energy from the joulemeter, rise from the thermometer, c = ΔE/(mΔθ); (c) one temperature everywhere, less energy escapes; (d) 42 000 J; (e) a large mass and a large rise store a lot of energy, far above 100 °C.

Watch for

“(a) 1000 cm³ of water has a mass of 1000 kg.” — Each cubic centimetre has a mass of 1 g: 1000 cm³ is 1000 g, which is 1 kg.
“(e) Bricks are used because they have a higher specific heat capacity than water.” — Water’s specific heat capacity is higher. Bricks win because a large mass of them can be heated to several hundred degrees, which water cannot be: it boils at 100 °C.