I need help with this word problem.

If a current of 6.3 amperes flow through an electric heater that has a resistance of 19 ohms, what is the voltage drop across the heater? If it costs $0.10 per kilowatt-hour for electrical energy, how much will it cost to operate the electric heater for eight hours? How much electric energy in joules was used to operate the electric heater for the eight hours?

Thanks!

This word problem, like most, is mostly a matter of picking the right equations.

The equations you need in this case are

V = I x R
P = I x V

P is power (in Watts), I is current (in Amperes), V is potential [sometimes called voltage] (in Volts), R is resistance (in Ohms).

Using the resistance and the current you're given, you can calculate the potential (voltage) drop using the first equation. With the potential (voltage) drop and the current you're given, you can use the second equation to calculate the power.

Kilowatt-hours are a unit of energy which you calculate, as the name suggests, by multiplying power (in kilowatts) by time (in hours). Since there are 1,000 Watts in a kilowatt, divide the power you calculated earlier by 1,000. Then, multiply this by the amount of time (8 hours, from the question) and you'll have a value in kilowatt-hours.

Then, you multiply the cost per kilowatt-hour by the number of kilowatt-hours to get the total cost.

Joules are also a unit of energy, so you also can calculate them by multiplying power and time together. In this case, however, you want everything in the standard units (time in seconds, power in watts), because Joules are one of the standard units. You already calculated the power in watts, so that's done, but you still need to calculate the number of seconds.

Take the number of hours (8) and multiply them by the number of minutes in an hour (60) to get the number of minutes in 8 hours. Then take that number and multiply it by the number of seconds in a minute (60) and you'll have the number of seconds in 8 hours.

Take the seconds and the watts, multiply them together, and you'll have the energy used in Joules.