2kg of water is heated with a heating coil which draws 3.5A from a 200V 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:

Option D = 10°

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