A heater coil connected to 200 V has a resistance of 80 Ω. If the heater is plugged in for the time t such that 1 kg of water at 20°C attains a temperature of 60°C. Find the power of the heater.

[19]
A heater coil connected to

200 V

has a resistance of

80 Ω

, when the heater is plugged in for the time

t

such that

1

kg

of water at

20 ° C

attains a temperature of

60 ° C

, the power of the heater will be 500W.
We know that the power of heater is

P = V^{2} / R

Here,

V

is the voltage,

P

is the power, and

R

is the resistance.

\Rightarrow P = {(200)}^{2} / 80 W

By simplifying, we get

P = 500 W

A heater coil connected to $200 V$ has a resistance of $80 Ω$ . If the heater is plugged in for the time $t$ such that $1$ $kg$ of water at $20 ° C$ attains a temperature of $60 ° C$ . Find the power of the heater.