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O-Level Thermal Properties of Matter
What the O-Level syllabus expects for Thermal Properties of Matter, and how to practise it.
What the syllabus expects
- Describe internal energy as a store equal to the total random-motion kinetic energy of the particles plus the total potential energy between them in the system.
- Define heat capacity and specific heat capacity.
- Describe melting/solidification and boiling/condensation as energy transfers that occur without any temperature change.
- Explain how boiling differs from evaporation.
- Define latent heat and specific latent heat.
- Explain latent heat by reference to how particles in a body behave.
- Draw and interpret a cooling curve.
How it's examined
Questions on this topic most often ask you to state, calculate, estimate, find. About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
The container holds 120 g of noodles at 28 °C. A quantity of 350 g of water at 85 °C is then poured in. Take the specific heat capacity of water as 4200 J / (kg°C) and that of the noodles as 2400 J / (kg°C). Assuming no heat escapes to the surroundings or the container, find the final temperature reached by the mixture.
Show the worked answer
Heat lost by water = heat gained by noodles (no losses). Let final temperature be T. m_w c_w (85 - T) = m_n c_n (T - 28). (0.350 x 4200)(85 - T) = (0.120 x 2400)(T - 28). 1470(85 - T) = 288(T - 28). 124950 - 1470T = 288T - 8064. 133014 = 1758T. T = 75.7 degrees C.
Example 2 (2 marks)
An electric shower warms water using a heating element inside a storage tank. The tank holds 100 kg of water and starts off completely full at 25 degrees C. Before showering, the water must be raised to 60 degrees C. The heating element is rated 230 V, 1.8 kW. (b) Calculate the quantity of heat energy needed to warm the full tank of water up to 60 degrees C. (Use a specific heat capacity of water of 4200 J kg^-1 K^-1) heat energy = ......
Show the worked answer
Q = mcΔT ΔT = 60 - 25 = 35 K Q = 100 x 4200 x 35 = 1.47 x 10⁷ J
Example 3 (3 marks)
A frozen soup, treated as pure water, starts at -18 C and must just reach liquid at 0 C. Estimate the shortest time a 750 W microwave oven needs to thaw 0.25 kg of it. Use 2100 J/(kg C) for ice's specific heat capacity and 340 000 J/kg for its specific latent heat of fusion.
Show the worked answer
Step 1 - identify the two separate energy transfers. The soup starts as ice at −18 °C and must end as liquid at 0 °C, so two things must happen in turn: (i) the ice is warmed from −18 °C up to its melting point, 0 °C, while still solid; (ii) the ice is then melted at 0 °C, with no further temperature rise. Step 2 - state the principle for each. Warming a substance without changing state: E = mcΔθ where m is the mass (kg), c the specific heat capacity (J/(kg °C)) and Δθ the temperature change (°C). Changing state at constant temperature: E = mL where L is the specific latent heat of fusion (J/kg). No Δθ appears, because the temperature does not change while melting. Step 3 - energy needed to warm the ice. Δθ = final temperature − initial temperature = 0 − (−18) = 18 °C E1 = mcΔθ E1 = 0.25 kg × 2100 J/(kg °C) × 18 °C E1 = 9450 J Step 4 - energy needed to melt the ice. E2 = mL E2 = 0.25 kg × 340 000 J/kg E2 = 85 000 J Step 5 - total energy required. E = E1 + E2 E = 9450 J + 85 000 J E = 94 450 J Step 6 - link energy to power and time. Power is the rate of energy transfer, P = E/t, so E = Pt. Asking for the SHORTEST time means assuming the oven is 100% efficient, that is every joule it delivers goes into the soup and none is lost. Rearranging for t: t = E/P Step 7 - substitute. t = 94 450 J / 750 W t = 125.9 s To two significant figures this is about 130 s, roughly two minutes.
More O-Level Pure Physics topics
Physical Quantities, Units and Measurement · Kinematics · Dynamics · Turning Effects of Forces · Pressure · Energy · all of O-Level Pure Physics