What is the efficiency of the installation if, at a current in the electric stove of 2.5 A and at a voltage of 220 V. 570 g of water in a saucepan is heated from 10 C to a boil in 12 minutes?
To find out the efficiency of the specified installation, we will use the formula (the heating of the pan is not taken into account): ηx = Sv * mw * (100 – t0) / (U * I * t).
Const: St. – beats. heat capacity of water (Sv = 4.18 * 10 ^ 3 J / (kg * K).
Data: mw is the mass of water in the pan (mw = 570 g; in the SI system mw = 0.57 kg); t0 – early. temperature (t0 = 10 ºС); U – mains voltage (U = 220 V); I is the current in the electric stove (I = 2.5 A); t is the time spent on heating (t = 12 min; in the SI system t = 720 s).
Let’s perform the calculation: ηx = Sv * mw * (100 – t0) / (U * I * t) = 4.18 * 10 ^ 3 * 0.57 * (100 – 10) / (220 * 2.5 * 720) = 0.5415 or 54.15%.
Answer: The efficiency of this installation is 54.15%.
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