Question

A clothes dryer has a power rating of $4500\;W$. How long did it take to dry a load of laundry if electric power costs $\$0.09\;kW/h$ and the cost of using the dryer is \[\$0.63\]

Answer 1

Hint: We know that power in physics refers to the amount of energy spent or transferred to do any work at any given time interval. Here, the electrical energy is converted to heat energy by the clothes dryer to dry the clothes. Also the power rating and the cost of the dryer is given, we need to calculate the time taken by the dryer.
Formula used:
$t=\dfrac{power\; consumed}{power\;rating}$ and $p=\dfrac{rate\; per\; unit}{total\;cost}$

Complete answer:
Generally, power rating of electrical equipment is the maximum power which can be allowed through that given equipment to do work. Here in the given we have a cloth dryer which converts the electrical energy into heat energy to dry the clothes which are added to the clothes dryer.
Here, it is given that the power rating of the dryer is $4500\;W$, that is we need, $4500\;W$ of electricity to drive the dryer for an hour, also the rate of the current consume is given as \[\$0.63\] . Here, the total cost of running the dryer is given as $\$0.09\;kW/h$ .
Then the time taken to run the dryer is given as $t=\dfrac{power\; consumed}{power\;rating}$. Where, power consumed is given as $p=\dfrac{rate\; per\; unit}{total\;cost}$
substituting we get
$\implies p=\dfrac{0.63}{0.09}=7kW$
$\implies t=\dfrac{7kW}{4.5KW}=1.56hrs$
Thus it takes around $1.56\;hrs$ for the dryer to dry the clothes with the given power rating.

Note:
The power rating is the guidelines maintained and followed by the manufactures to protect the equipment from electrical damages, such as short circuit or electrical fire. In other words, we can also call the power rating a safety margin used by the electrical industries to avoid accidents.