2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water? [specific heat capacity of water = 4200 jkg -1k-1]
A. 30° B. 25° C. 15° D. 10°
Correct Answer: D
Explanation
Electricity heat energy supplies by the heater = lvt = 3.5 x 200 x (2 x 60) = 84000j. and this heat is supplied to the 2 kg of the water MC ∆ t = 84000 2 x 4200 x ∆ t = 84000 ∴ ∆t = (84000)/8400 = 10°c ∴ increase in temp. of water = 10°c
