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]
- 30°
- 25°
- 15°
-
10°
The correct answer is: 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