Question

A \[500W\] heater is designed to operate from a \[200V\] line. By what percentage will its heat output drop its line voltage drop to \[160V\] ? Find the heat produced in \[10minutes\] .

Answer 1

Hint:In order to this question, first we will rewrite the given facts of both cases and then find the power generated in both cases separately. And then we will find the exact value of powers in both cases. And finally we can find the heat produced in $10\min $ .

Complete step-by-step solution:
As we can see, the question is given in 2 phases. So, we will first rewrite the facts of 1st case-
In 1st case-
Power of heater is, ${P_1} = 500W$
And, the heater operates from voltage, ${V_1} = 220v$
So, we will apply the formula of Power in terms of Resistance:
$\because {P_1} = \dfrac{{{V_1}^2}}{R} = \dfrac{{{{(220)}^2}}}{R}$ …………(i)
Similarly, in 2nd case-
The voltage is, ${V_2} = 160v$
And, the power is, ${P_2} = ?$
So, again we will apply the formula of Power in terms of Resistance:
$\because {P_2} = \dfrac{{{V_2}^2}}{R} = \dfrac{{{{(160)}^2}}}{R}$ ……….(ii)
As we can see, Resistance can’t be changed.
Now, we will divide Eq(i) by eq(ii), we get-
$\therefore \dfrac{{{P_1}}}{{{P_2}}} = {(\dfrac{{200}}{{160}})^2} = {(\dfrac{5}{4})^2} = \dfrac{{25}}{{16}}$
$ \Rightarrow {P_2} = \dfrac{{16}}{{25}} \times {P_1}$ ……..(iii)
So, we can see that the heat is also dropping.
Now, we will find the change in power percentage:
$
  \therefore \Delta \% P = \dfrac{{{P_1} - {P_2}}}{{{P_1}}} \times 100 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{P_1} - \dfrac{{16}}{{25}}{P_1}}}{{{P_1}}} \times 100 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{25 - 16}}{{25}} \times 100 = 36\% \\
 $
As, ${P_1}$ is given i.e.. $500W$ .
So, in equation(iii):-
$
  \therefore {P_2} = \dfrac{{16}}{{25}} \times {P_1} \\
  \,\,\,\,\,\,\,\,\,\, = \dfrac{{16}}{{25}} \times 500 = 320W \\
 $
Therefore, the power generated by the heater in the 2nd case when it is dropping the voltage is $320W$ .
Now, we can find the heat produced it in $10\min $ :
$
  \therefore H = (10 \times 60s) \times 320 \\
  \,\,\,\,\,\,\,\,\, = 192 KJ \\
 $
Hence, the heat produced by a heater in $10\min $ is $192KJ$ .

Note:Heat is a type of energy that can be used to generate work. Units of work are used to measure the amount of energy. It's measured in joules, pounds, kilowatt-hours, or calories. Heat is a kind of energy that can be turned into other forms. Heat is transformed to mechanical energy in motorised vehicles, for example.