Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is

  • Option 1)

    50 s

  • Option 2)

    100 s

  • Option 3)

    150 s

  • Option 4)

    200 s.

 

Answers (2)

As we learnt in

Condition for Maximum Power -

R= \frac{nr}{m}

P_{max}=(mn)\frac{E^{2}}{4r}

 

- wherein

mn - total number of cells .

 

 Electrical energy converted into heat energy.

836\times t=1000\times 1\times \left ( 40-10 \right )\times \left ( 4.18 \right )

t = \frac{1000\times 30\times 4.18}{836}

= 150 sec

 


Option 1)

50 s

This option is incorrect.

Option 2)

100 s

This option is incorrect.

Option 3)

150 s

This option is correct.

Option 4)

200 s.

This option is incorrect.

N neha

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