Question

13) An electric iron of resistance 50 ohm generates 378 k.cal of heat in 30 minutes.
What is the voltage of the mains ?​

Answer 1

Answer:

Given:

Resistance of iron = 50 ohm

Heat generated = 378.6 kCal

Time taken = 30 minutes

To find:

Voltage of the mains

Calculation:

First we will try to convert the heat generated from kCal unit to Joules unit.

$378 \: kcal = 378 \times {10}^{3} cal$

We know that :

$1 \: cal = 4.2 \: joule$

$= > 378 \times {10}^{3} cal = 1587.6 \times {10}^{3} joule$

Now we will find out the power of the iron

$power = \dfrac{heat}{time}$

$= > power = \dfrac{(1587.6 \times {10}^{3} )}{(30 \times 60)}$

$= > power = 882 \: watt$

We know the relationship between power , resistance and Voltage as follows :

$power = \dfrac{ {(voltage)}^{2} }{resistance}$

$= > 882 = \dfrac{ {v}^{2} }{50}$

$= > {v}^{2} = 882 \times 50$

$= > {v}^{2} = 44100$

$= > v = \sqrt{44100}$

$= > v = 210 \: volt$

So final answer :

$\boxed{ \huge{ \blue{ \bold{v = 210 \: volt}}}}$

Answer 2

$\large \underline{ \red{ \boxed{ \bf \orange{Required \: Answer}}}}$