Solids, Liquids and Gases
4 marks · EasyIn an experiment, a student heats a 0.25 kg block of copper from 18 °C to 38 °C. The specific heat capacity of copper is 390 J/kg °C.
State the equation used to calculate the change in thermal energy.
Calculate the change in thermal energy of the copper block as it is heated.
State what happens to the thermal energy supplied when the temperature of the copper block stops rising.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) ΔQ = m × c × Δθ
(b) 1,950 J
(c) It is transferred to the surroundings or the block loses energy to the environment.
Common mistakes: using wrong equation; not subtracting the initial from final temperature
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Gives the correct equation for thermal energy: ΔQ = m × c × Δθ (in symbols or words).
M1 (b) · 1 mark · Substitutes correct values into the formula.
A1 (b) · 1 mark · Correct calculation of ΔQ = 0.25 × 390 × 20 = 1,950 J (accept answers between 1,900 and 2,000 J). (follow-through)
B2 (c) · 1 mark · Explains energy is transferred to the surroundings or lost (or heating stops so no more temperature rise).
Worked solution
(a) The equation used: ΔQ = m × c × Δθ
(b) Substituting the values: m = 0.25 kg, c = 390 J/kg °C, Δθ = 38 − 18 = 20 °C
(c) When the block stops heating, the thermal energy is lost to the surroundings or the block is in thermal equilibrium.