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Question

An electric heater having a heating coil of 150 ohms connected with a supply voltage of 220 V is used to heat water. The time taken to increase the temperature of 1 kg of water by 20ºC to 60ºC is


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Solution

Step 1: Given data

  1. The resistance of the coil is, R=150Ω
  2. The mass of the water is, m=1kg
  3. Supply voltage is V=220volt
  4. The increase in temperature is T=60-20ºC=40ºC.

Step 2: Joule's law of heating and heat loss

  1. Joule's law states that when a current I is flowing in a conductor of resistance R for a time T sec, then the produced heat on the conductor is directly proportional to the square of the electric current.
  2. Joule's law is defined by the form, H=I2RT , where, Q is the produced heat on the conductor.
  3. Heat loss by a conductor is defined by the form, H=msT, where, m is the mass of the conductor, s is the specific heat, and T is the temperature difference.

Step 3: Calculation of produced heat

Now applying Joule's law, heat generated by the conductor,

H=I2RT=V2R×T(V=IR)orH=2202150×T=48400150=484015TorH==484015T..........(1)

Step 4: Calculation of heat loss

Again, heat loss,

H=msT=1×4200×60-20orH=4200×40orH=168000.................(2) (specific heat of water is s=4200J/kg°C )

Step 5: Calculation of time

Using the property of heat and calorimetry
heat loss = heat gain. So, from equations 1 and 2

484015×T=168000orT=1680004840×15=520.67orT=520.67seconds.

Therefore, the time taken to increase the temperature of 1 kg of water from 20ºC to 60ºC is 520.67seconds.


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