When the heating was turned on, the air in a room with a volume of 20 meters in a cube warmed up from 10 to 20 degrees. What mass of water has passed through the heating batteries if the water in the batteries cools down from 60 to 58 degrees?
Given: V (air volume) = 20 m3; tв1 (initial air temp.) = 10 ºС; tв2 (end air temp.) = 20 ºС; tb1 (initial temp. battery) = 60 ºС; tb2 (battery end temp.) = 58 ºС.
Constants: ρ (air density) = 1.225 kg / m3; C1 (heat capacity of air) = 1000 J / (kg * K); C2 (heat capacity of water) = 4200 J / (kg * K).
Let us express the mass of water: C1 * ρ * V * (tv2 – tv1) = C2 * m2 * (tb1 – tb2); m2 = C1 * ρ * V * (tv2 – tv1) / (C2 * (tb1 – tb2)).
Calculation: m2 = 1000 * 1.225 * 20 * (20 – 10) / (4200 * (60 – 58)) = 29.17 kg.
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