An aluminum kettle weighing 300 g heats up 2 liters of water from 16 to 100

An aluminum kettle weighing 300 g heats up 2 liters of water from 16 to 100 degrees Celsius. What amount of heat is spent on heating water?  

Problem data: m1 (kettle weight) = 300 g (0.3 kg); V2 (water volume) = 2 l (2 * 10-3 m3); t1 (starting temp. kettle, water) = 16 ºС; t2 (final heating temp.) = 100 ºС.

Constants: Ca (specific heat capacity of aluminum) = 920 J / (kg * K); ρ (water density) = 1000 kg / m3; Sv (specific heat capacity of water) = 4200 J / (kg * K).

1) Teapot heating: Q1 = Ca * m1 * (t2 – t1) = 920 * 0.3 * (100 – 16) = 23 184 J.

2) Water heating: Q2 = Sv * m2 * (t2 – t1) = Sv * ρ * V2 * (t2 – t1) = 4200 * 1000 * 2 * 10-3 * (100 – 16) = 705 600 J.