What is the time required to heat the water in the following case?
A kettle has power 2.5 kW. A liter of water is poured into it at 20 C and the kettle is switched on.
a) How long will it take the kettle to boil?
b) If the kettle fails to switch on, how long will it take for all the water to evaporate?
Note: Any heat loss may be ignored.
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You can only ask one question at a time. So, I have deleted all but one.
Now the kettle can deliver 2.5 kW of power. One liter of water is poured into it. The specific heat of water is 4186 J/(kg*K).
So it takes 4.186 kJ to raise one kilogram of water by one K. The mass of one liter of water is approximately 1 kg.
To bring the water to boiling it has to be raised to 100 C, the initial temperature is 20 C.
Therefore the energy required is (100 - 20)*4.186 = 334.88 kJ.
The power rating of the kettle is 2.5 kW. So to deliver 334.88 kJ the time required is 334.88/ 2.5 = 133.95 s.
a) The required time is 133.95 s.
b) If the kettle does not switch on, the water will not evaporate.
a) you also need to calculate Latent Heat of evaporation...
b) depends on variables such as the ambient air temperature, air movement and relative humidity of the air around the kettle. And dew point.
For example, if the ambient air is at 95 degrees C and the humidity level is low, then the water will most likely evaporate within a few hours, perhaps even a few minutes.
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