GCSE level Physics exam revision notes on specific heat

Thermal energy Part 2.2 Practice exam calculation questions based on specific heat data of materials or calculating specific heat capacity from given information

Q1 A 0.5 kg block of copper absorbed 1520 J of energy and its temperature rose by 8.0^oC.

Calculate the specific heat capacity of copper.

∆E = m x c x ∆T

SHC_copper = ∆E / (m x ∆T)

SHC_copper = ∆E / (m x ∆T) = 1520 / (0.5 x 8.0) = 380 J/kg^oC

 

Q2 A block bronze has a specific heat capacity of 400 J/kg^oC.

If a 1500 g block of bronze absorbs 3000 J of energy, what would be the rise in its temperature?

∆E = m x c x ∆T,  1500 g = 1.500 kg

so rearranging: ∆T = ∆E / (m x SHC_bronze) = 3000 / (1.5 x 400) = 5.0^oC

Q3 Calculate the thermal energy in kJ required to heat 2 kg of water (SHC = 4180 J/kg^oC) from 20^oC to 100^oC.

∆E = m x c x ∆T = 2 x 4180 x (100-20) = 6.68 x 10⁵ J

Thermal energy required = 668 kJ

 

Q4 A storage tank of water contains 500 kg of hot water at 80^oC. (SHC water = 4180 J/kg^oC)

How many MJ of thermal energy can be transferred to a heating system if the temperature of the water falls to 25^oC?

∆E = m x c x ∆T = 500 x 4180 x (80-25) = 1.15 x 10⁸ J

Thermal energy transferred = 1.15 x 10⁸ / 10⁶ = 115 MJ (1.15 x 10² MJ, 3 sf)

 

Q5 A more complicated problem to solve - so 'double' think carefully!

Suppose 2 kg of steel at a temperature of 80^oC is placed in 10 kg (~10 litres) of water at 15^oC.

Calculate the final temperature of the materials.

SHC steel = 450 J/kg^oC,  SHC water = 4180 J/kg^oC

The loss in J from the steel energy store = gain in J of the thermal energy store of the water.

At 'thermal equilibrium' they will be both at the same temperature, lets call this T.

The temperature fall of the steel is (80 - T), the temperature rise of the water is (T - 15)

The loss from the steel thermal energy store = ∆E = m x c x ∆T = 2 x 450 x (80 - T) = 900 x (80 - T) J

The gain of the water thermal energy store  = ∆E = m x c x ∆T = 10 x 4180 x (T - 15) = 41800 x (T - 15) J

So, change in steel thermal energy store = change in water thermal energy store

 900 x (80 - T) = 41800 x (T - 15)

72000 - 900T = 41800T - 627000

(changing sides, changing signs, we get)

72000 + 627000 = 900T + 41800T

699000 = 42700T

T = 699000 / 42700 = 16.4^oC (3 sf)

Q6 A night storage heater contains 75 kg of concrete (SHC = 800 J/kg^oC)

At night when electricity is cheaper and the house cooler, how many kJ of thermal energy is needed to raise the temperature of the concrete from 8^oC to 35^oC.

∆E = m x c x ∆T = 75 x 800 x (35 - 8) = 1.62 x 10⁶ J

1.62 x 10⁶ / 1000 = 1620 J

 

Q7 This question is based on a physics experiment I did in school back in circa 1961.

Its a good little experiment to do.

Accurately weigh 100.0 g (~100 ml) of water into a beaker at room temperature.

A 50.0 g brass weight was held by a thin wire string or with a pair tongs in the centre of a roaring Bunsen flame.

After a few minutes the brass weight was plunged into a beaker of 100 g cold water whose initial temperature was 19.5^oC.

After gently stirring carefully with a thermometer the temperature of the water rose to 55.5^oC.

SHC of brass = 380 J/kg^oC, SHC water = 4180 J/kg^oC

(a) If the flame temperature is T, calculate the two thermal energy store transfers.

Thermal energy transfer from brass weight = ∆E = m x c x ∆T = (50/1000) x 380 x (T - 19.5) = 19.0 (T - 19.5) J

Thermal energy transfer from brass weight = ∆E = m x c x ∆T = (100/1000) x 4180 x (55.5 - 19.5) = 15048 J

(b) Calculate the flame temperature T

Thermal energy transfer from brass weight (J) = thermal energy transfer to water (J)

19.0 (T - 19.5) = 15048

19.0T - 370.5 = 15048

19.0T = 15418.5

T = 15418.5 / 19.0 = 812^oC (3 sf)

(b) Suggest some sources of error in the experiment

(i) the weight loses some thermal energy to air on transfer - possibly quite a bit, since the experimental answer you get is often well below the real answer for the flame temperature of a bunsen flame ~1000-1200^oC.

(ii) the glass beaker absorbs some heat, therefore its heat capacity has been ignored

(iii) the lower part of the wire will transfer a little heat - but if you use brass wire, you could weigh that too and make sure everything is immersed in the water!

Q8 A car of mass 1000 kg is moving at 20.0 m/s is brought to a halt by sharp braking.

Assume all the kinetic energy is converted to thermal energy by four steel brake drums.

If the total mass of the steel drums of the brakes is 25.0 kg, assuming no heat losses, calculate the maximum rise in temperature the brake drums will attain.

(a) First calculate the kinetic energy (KE) of the car.

KE = ½ mv²

KE = ½ x 1000 x 20² = 2.0 x 10⁵ J

(b) Calculate the temperature rise of the braking system (SHC of steel = 450 J/kg^oC)

∆E = m x c x ∆T

∆T = ∆E / (m x c) = 2.0 x 10⁵ / (25 x 450) = 17.8^oC (3sf)

The next two questions are more tricky and involve using electricity formulae: Electricity power calculations, P = IV

The specific heat capacity of water is 4180 J/kg^oC. Be careful with all the units.

Q9 A small electric kettle is connected to the 240 V AC mains supply and uses a current of 8.0 A.

It contains 1.5 kg of water (~1.5 litres, 1500 ml, 1500 cm³) at 20^oC.

(a) Calculate the power of the kettle and the rate of energy transfer.

P (W) = I (a) x V (p.d. in volts) = 8 x 240 = 1920 W = 1920 J/s

(b) If the kettle is switched on for 2.0 minutes, how much energy is transferred to the thermal energy store of the water?

P = 1920 J/s, energy transferred = power x time = 1920 x 2 x 60 = 230400 J = 2.3 x 10⁵ J

(c) What temperature will the hot water rise too?

∆E = mass of water x SHC_H2O x temperature rise

∆E (J) = m (kg) x c (J/kgoC) x ∆T (^oC)

Rearranging: ∆T = ∆E / (m x SHC_H2O)

∆T = 230400 / (1.5 x 4180) = 36.7 ^oC

Therefore temperature after 2 minutes of heating is 20 + 36.7 = 56.7^oC

(d) How much energy is needed to raise the water temperature from 20^oC to 100^oC?

∆E = mass of water x SHC_H2O x temperature rise

∆E = 1.5 x 4180 x (100 - 20) = 5.016 x 10⁵ J = 5.01 x 10⁵ J (3 sf)

(e) Starting with the cold water at 20^oC, how long would it take to boil the kettle' in minutes and seconds?

What assumptions have you made for this calculation?

The kettle's power s 1920 W, that is 1920 J/s, and energy needed =  5.016 x 10⁵ J

P = E / t, so t = E / P = 5.016 x 10⁵ / 1920 = 261.25 s.

Time to boil = 4 mins and 21 seconds (to the nearest second).

This calculation assumes all the electrical energy converted to thermal energy, actually increased the thermal energy store of the water.

However, you would always get a small amount of wasted thermal energy transferring to the thermal energy store of the kettle casing by conduction, and convection and radiation transfer to the thermal energy store of he surrounding air.

 

Q10

A hot water cylinder (immersion heater) is quoted as having a capacity of '120 litres of water' (120 kg of water).

The tank is fitted with a 3 kW heating element working off 240 V mains electricity.

(a) How much energy is needed to heat fresh cold water at 12^oC to a temperature of 70^oC?

∆E = mass of water x SHC_H2O x temperature rise

Energy needed = 120 x 4180 x (70 - 12) = 29 092 800 J = 2.91 x 10⁷ J (3 sf)

(b) How long will it take to heat the water up? (to the nearest minute)

P (J/s) = E (J) / t (s), 3000 = 29 092 800 / t

time = E / P = 29 092 800 / 3000 = 9697.6 seconds

9697.6 / 60 = 161.6.  time required = 162 minutes (sf)

(c) What current is flowing through the heating element?

P = I x V,  I = P / V = 3000 / 240 = 12.5 A

(d) What is the resistance of the heating element?

V = IR,  R = V / I = 240 / 12.5 = 19.2 Ω

(e) If the immersion was fitted with a 6 kW heating element, how would this affect the time to heat water up to a certain temperature?

This is twice the power of the 3 kW heating element.

P = E / t,  t = E / P, for the same quantity of water and the same temperature rise, it would take half the time compared to the 3 kW heater.