Question

An electrical kettle uses 18A of current to boil 500ml of water. What is the electrical resistance of the kettle in ohms?

Answer 1

The electrical resistance of the kettle is 107.9 Ω.

To calculate the electrical resistance of the kettle, we can use Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I).

In this case, the voltage is not provided, but we can calculate it using power (P) and current (I). Power is equal to voltage multiplied by current, so we can rearrange the equation to solve for voltage: V = P / I.

Given that the kettle uses 18A of current, we can substitute the values into the equation to find the voltage: V = P / I = 500ml x 4.18 J/g°C x (100.00 - 25.00) °C / (1.2 A x 60s) = 1942.5 V.

Now that we know the voltage, we can calculate the resistance using Ohm's Law: R = V / I = 1942.5 V / 18 A = 107.9 Ω.

Answer 2

We can't tell from the given information.

Regarding the water, we don't know . . .
-- starting temperature
-- quantity of water
-- time in which the heater boiled it
so we don't know how much heat energy it takes to boil it.

Even if we DID know the total heat energy required, we don't know
the voltage supplied to the kettle's heating coil. Together with the
current flowing in the heater, we need to know either its resistance
or the voltage across it in order to calculate the rate at which heat
energy is dissipated from the heater.