What mass of water can be heated from 20 degrees Celsius to 80 C due to the energy consumed in 10 minutes by kettles connected to a 220 V network. With a current of 2 A, the KPT of the kettle is 80%? Specific heat capacity of water 4200J / kg s
t1 = 20 ° C.
t2 = 80 ° C.
C = 4200 J / kg * ° C.
T = 10 min = 600 s.
U = 220 V.
I = 2 A.
Efficiency = 80%
Efficiency = 80% of an electric kettle shows that 80% of the heat energy that is released in the spiral of the kettle Qh goes to heating the water Qw: Efficiency = Qw * 100% / Qh.
The heat released in the teapot Qh can be expressed by the formula: Qh = I * U * T.
The heat required for heating water Qw is expressed by the formula: Qw = C * m * (t2 – t1).
Efficiency = C * m * (t2 – t1) * 100% / I * U * T.
The mass of heated water in the teapot m will be determined by the formula: m = efficiency * I * U * T / C * (t2 – t1) * 100%.
m = 80% * 2 A * 220 V * 600 s / 4200 J / kg * ° C * (80 ° C – 20 ° C) * 100% = 0.84 kg.
Answer: the kettle for the specified time can heat m = 0.84 kg of water.
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