In an aluminum saucepan weighing 0.5 kg, 3 kg of water is heated from 10 ° to 60 °. How much heat is needed for this?

Given: m1 = 0.5 kg, m2 = 3 kg, t1 = 10 ° C, t2 = 60 ° C, c1 = 920 J / kg • ° C, c2 = 4200 J / kg • ° C
Find: Q
Solution: You need to heat both the pot and the water from the same temperature to another again at the same temperature. This can be done using the formula: Q = cm (t2-t1)
1) heat the pan: Q1 = c1 • m1 • (t2-t1)
2) we heat the water: Q2 = c2 • m2 • (t2-t1)
3) add all the energy received from the pan and water together: Q = Q1 + Q2 = c1 • m1 • (t2-t1) + c2 • m2 • (t2-t1), now we can move the temperature difference forward: Q = (t2- t1) (c1 • m1 + c2 • m2), now we substitute the values and count: Q = (60-10) (920 • 0.5 + 4200 • 3) = 50 • (460 + 12600) = 50 • 13060 = 653000 J = 653 kJ
Answer: 653 kJ