How much heat is needed to heat the volume v = 1 liter of water in a steel kettle weighing m = 1 kg from
| May 7, 2021 | education
How much heat is needed to heat the volume v = 1 liter of water in a steel kettle weighing m = 1 kg from temperature t1 = 20 degrees Celsius to temperature t2 = 80 degrees Celsius
Data: V1 (water volume) = 1 l (0.001 m3); m2 (kettle) = 1 kg; t1 (initial temp.) = 20 ºС; t2 (end temp.) = 80 ºС.
Constants: ρ1 (water density) = 1000 kg / m3; C1 (specific heat capacity of water) = 4200 J / (kg * ºС); С2 (specific heat capacity of steel) = 500 J / (kg * ºС).
1) Mass of water: m1 = ρ1 * V1 = 0.001 * 1000 = 1 kg.
2) Water heating: Q1 = C1 * m1 * (t2 – t1) = 4200 * 1 * (80 – 20) = 252000 J.
3) Teapot heating: Q2 = C2 * m2 * (t2 – t1) = 500 * 1 * (80 – 20) = 30,000 J.
4) Q = 252000 + 30000 = 282000 J (282 kJ).
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