In a vessel with kerosene, the mass of which is 100 g, at a temperature of 20 ° C, a steel part weighing 10 g was placed at a temperature of -20 ° C. Find the steady-state temperature.
Data: m1 (mass of kerosene) = 100 g = 0.1 kg; t1 (initial temperature of kerosene) = 20 ºС; m2 (weight of the steel part) = 10 g = 0.01 kg; t2 (initial part temperature) = -20 ºС.
Reference data: C1 (specific heat capacity of kerosene) = 2100 J / (kg * K); C2 (specific heat capacity of steel) = 500 J / (kg * K).
The steady-state temperature is determined from the equality: C1 * m1 * (t1 – t3) = C2 * m2 * (t3 – t2).
2100 * 0.1 * (20 – t3) = 500 * 0.01 * (t3 – (-20)).
4200 – 210t3 = 5t3 + 100.
215t3 = 4100.
t3 = 4100/215 ≈ 19 ºС.
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