Home
Class 12
PHYSICS
An electric heater of resistance 10Omega...

An electric heater of resistance `10Omega` connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from `30^(@)C` to `60^(@)C`. (The specific heat of water is `s=4200J kg^(-1)`/C`)

Text Solution

Verified by Experts

According to Joule.s heating law `H=I^(2)Rt`
The current passed through the electrical heater `=(220V)/(10Omega)=22A`
The heat produced in one second by the electrical heater `=(220V)/(10Omega)=22A`
The heat produced in one second by the electrical heater `H=I^(2)R`
The heat produced in one second `H=(22)^(2)xx10=4840J=4.84kJ`. In fact the power rating of this electrical heater is 4.84 k W.
The amount of energy to increase the temperature of 1 kg water from `30^(@)C` to `60^(@)C` is
`Q=ms DeltaT`
Here `m=1kg, s=4200J kg^(-1), DeltaT=30`
so `Q=1xx4200xx30=126kJ`
The time required to produce this heat energy
`t=(Q)/(I^(2)R)=(126xx10^(3))/(4840)=26.03s`
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    FULL MARKS|Exercise TEXTBOOK EVALUATION (Choose the correct answer)|15 Videos

  • CURRENT ELECTRICITY

    FULL MARKS|Exercise TEXTBOOK EVALUATION (Short Answer Questions)|21 Videos

  • COMMUNICATION SYSTEMS

    FULL MARKS|Exercise ADDITIONAL QUESTIONS  (Additional problems)|3 Videos

  • DUAL NATURE OF RADIATION AND MATTER

    FULL MARKS|Exercise ADDITIONAL QUESTIONS -( ADDITIONAL NUMERICAL PROBLEMS : )|10 Videos

Similar Questions

Explore conceptually related problems

Find the increase in mass when 1 kg of water is heated from 0^@C to 100^@C . Specific heat capacity of water = 4200 J kg^(-1) K^(-1) .

Calculate the heat energy required to raise the temperature of 2 kg of water from 10^@C to 50^@C . Specific heat capacity of water is 4200 JKg^(-1) K^(-1)

A 50 kg man is running at a speed of 18 kmh^(-1) If all the kinetic energy of the man can be used to increase the temperature of water from 20^(@)C to 30^(@)C .How much water can be heated with this energy?

What is the rice in temperature of 5 kg of water if it is given 84,000 J heat energy? Specific heat capacity of water =4200 J kg ^-1^@C^-1 .

An energy of 84000 J is required to raise the temperature of 2 kg of water from 60^@ C " to " 70^@ C . Calculate the specific heat capacity of water.

Heat needed to raise the temperature of 1 kg of water from 14.5^(@)C to 15.5^(@) C is ________ .

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates .Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10kg water and 0.2g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5^(@)C . Specific heat capacity of water = 4200J kg^(-1).^@C^(-1) and latent heat of vaporization of water = 2.27 xx 10^(6)J kg^(-1)

If 5 L of water at 50^(@)C is mixed with 4L of water at 30^(@)C , what will be the final temperature of water? Take the specific heat capacity of water as 4184 J kg^(-1) K^(-1)