Hence, the time required is approximately equal to 4 minutes and 39 seconds.
Given, Power, P = 300 W= 300J/s
Power= Energy per second
Mass, m= 0.25 kg
Initial Temperature= 20.0°C
Final Temperature= 100.0°C
Temperature difference, T =( 100-20)°C= 80.0°C
Specific heat of water, S= 4184 J/kg°C
Energy can be represented as
E=mST
where E is the energy, m is the mass, S is the specific heat of water and T is the temperature difference.
As Energy can be written as the product of time and power, E=Pt
Pt=mST
So,
t=(msT)/P
t=(0.25×4184×80)/300
Time, t = 278.934 seconds= 4 minutes and 39 seconds.
Hence, 4 minutes and 39 seconds will be required for a 300.0 W immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C.
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