What is the change in temperature of 2kg of water heated by a kettle using a voltage of 230V at 0.5A of current for 10 seconds? Assume no heat losses.
specific heat capacity of water, c = 4200J/kgK
First student should understand that due to no heat losses the energy provided by mains socket to kettle is the same energy as the heat which changes the water temperature. Q = mcdT = E = P x t
First electrical energy is calculated:E = P * t = I * V * tE = 230 * 0.5 * 10E = 1150 J
Application of energy conservation:E = Q = 1150JQ = mcdT This needs to be rearranged so that dT is the subject. dT = Q/(mc)Therefore our value for the temperature change is :dT = 1150/(24200)dT = 0.136K Which is as expected due to the kettle only being heated for 10 seconds!
WS
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