An aluminum container of mass 0.15 kg contains 0.25 kg of water. The specific heat capacity of aluminum is 900 J/(kg ∘C)900\text{ J}/(\text{kg }^\circ\text{C})900 J/(kg ∘C) and the specific heat capacity of water is 4200 J/(kg ∘C)4200\text{ J}/(\text{kg }^\circ\text{C})4200 J/(kg ∘C). How much energy, in joules, is needed to raise the temperature of the aluminum container and the water by 15∘C15^\circ\text{C}15∘C?
(0.15×900×15)−(0.25×4200×15)(0.15 \times 900 \times 15) - (0.25 \times 4200 \times 15)(0.15×900×15)−(0.25×4200×15)
(0.15×900×15)+(0.25×4200×15)(0.15 \times 900 \times 15) + (0.25 \times 4200 \times 15)(0.15×900×15)+(0.25×4200×15)
(0.15+0.25)×(900+42002)×15(0.15 + 0.25) \times \left(\frac{900 + 4200}{2}\right) \times 15(0.15+0.25)×(2900+4200)×15
(0.15+0.25)×(900+4200)×15(0.15 + 0.25) \times (900 + 4200) \times 15(0.15+0.25)×(900+4200)×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.