An electric heater is connected to a power supply with a voltage of 220 V, and the current passing through the electric heater is 5 a

An electric heater is connected to a power supply with a voltage of 220 V, and the current passing through the electric heater is 5 a

Power: P = UI = 220V × 5A = 1100W, so the heat generated is q = Pt = 1100W × 3600s = 3.96 × 106j. A: the heat generated through one hour is 3.96 × 106j

How much heat is generated in 10s when the "220 V, 50 W" electric furnace is connected to the lighting circuit? If it is connected to the 110 V power supply, how much heat is generated in 10s

1、W=Pt=50W*10s=500J
2. R = u ^ 2 / P = (220 V) ^ 2 / 50 W = 968 Ω
P '= u' ^ 2 / r = (110V) ^ 2 / 968 Ω = 12.5W
W'=P't=12.5W*10s=125J

In an enclosed space, will the light energy of an electric lamp eventually be converted into heat energy, so that the heat generated by the electric lamp is equal to that generated by an electric heater of the same power?

Yes, in general enclosed space, there will be no photoelectric reaction and photochemical reaction, and the light energy of electric lamp will eventually be converted into heat energy

In order to double the heat generated by the normal working electric heater in unit time and keep the power supply voltage unchanged, which of the following measures is feasible
A cut off half of the heating wire
B parallel a same resistance wire
Why can't a?
The answer on the first floor is not right. The resistance is half of the original, and the current has changed?

Because to make it work normally, it is not allowed to exceed the rated voltage and current. When the voltage is reduced by half, the constant voltage and the doubled current may exceed the rated range and burn out the electric heater