Q&A - Ask Doubts and Get Answers
Q

I need help with - Current Electricity - JEE Main

One kg of water, at 20^{\circ}C, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20\Omega. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to :

[ Specific heat of water =4200J/(kg^{\circ}C), Latent heat of water = 2260 kJ/kg]

 

  • Option 1)

    16 minutes 

  • Option 2)

    22 minutes 

  • Option 3)

    3 minutes 

  • Option 4)

    10 minutes 

 
Answers (1)
Views

 

Heat developed in a resistor -

When a steady current flows through a resistance R for time t , the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and H=i^2Rt

The power devoleped = \frac{energy}{time}=i^2R=iR=\frac{V^2}{R}      (from Ohm's law)

- wherein

Unit of heat is joule (J)

Unit of power is watt (W)

 

 

ms\Delta T+mL=\frac{V^{2}}{R}t

1\times 4200\times (100^{\circ}-20^{\circ})c+1\times 22.6\times 10^{5}=\frac{(200)^{2}}{20}t

336\times 10^{3}+226\times 10^{3}=2\times 10^{3}t

t=1298 sec. = 21.6 min \approx 22 minutes

 


Option 1)

16 minutes 

Option 2)

22 minutes 

Option 3)

3 minutes 

Option 4)

10 minutes 

Exams
Articles
Questions