An electric kettle is filled with 1.50 kg of water at 20C.The power of the kettle's element is 2.1 kW. After switching on, the water reaches boiling point in 240 s. Assume that all the energy produced by the element is transferred to the water. Calculate a value for the specific heat capacity of water.
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The energy(Q) absorbed by water is given by the equation;
`Q = mCtheta` where;
m = mass of water
C = specific heat capacity of water
theta = temperature change
This heat is gained by the electric energy of the kettle. Since there is no heat loss all the energy produced by kettle will be consumed by water to increase temperature.
Electric heat generated by kettle `= 2.1xx10^3xx240`
Heat absorbed by water `= 1.5xxCxx(100-20)`
`1.5xxCxx(100-20) = 2.1xx10^3xx240`
`C = 4200`
So the specific heat capacity of water is 4200J/(gkC)
Water boiled at 100C.
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