An aluminium container of mass 0.15 kg contains 0.25 kg of ethanol.
The specific heat capacity of aluminium is 900 J/(kg ∘C)900\text{ J}/(\text{kg }^\circ\text{C})900 J/(kg ∘C) and the specific heat capacity of ethanol is 2440 J/(kg ∘C)2440\text{ J}/(\text{kg }^\circ\text{C})2440 J/(kg ∘C).
How much energy, in joules, is needed to raise the temperature of the aluminium container and the ethanol by 15∘C15^\circ\text{C}15∘C?
(0.15×900×15)−(0.25×2440×15)(0.15 \times 900 \times 15) - (0.25 \times 2440 \times 15)(0.15×900×15)−(0.25×2440×15)
(0.15×900×15)+(0.25×2440×15)(0.15 \times 900 \times 15) + (0.25 \times 2440 \times 15)(0.15×900×15)+(0.25×2440×15)
(0.25+0.15)×(2440+9002)×15(0.25 + 0.15) \times \left(\frac{2440 + 900}{2}\right) \times 15(0.25+0.15)×(22440+900)×15
(0.25+0.15)×(2440+900)×15(0.25 + 0.15) \times (2440 + 900) \times 15(0.25+0.15)×(2440+900)×15
119 exam-style questions on CIE IGCSE Physics 2.2.2a Internal energy and temperature. Each one has a worked solution and a mark scheme showing where the marks go.