Question

an iron pot of mass 3.2kg containing 2 liters of water sits on top of an electric stove. The room temperature is 20C. The stove is turned on and over a period of 40 mins the lot and the water attain a temperature of 80C

A. What is the energy transferred to the pot during the time?

B. What is the energy transferred to the water during the time?

C. Assuming that none of then energy from the stove is lost, what is the power of the electric stove?

Answer 1

using equation Q = mSΔT

A. mass of iron = 3.2 Kg , S(iron) = 460.55 J/Kg-K, ΔT = 60K

Q( heat transferred to pot ) = 3.2 * 460.55 * 60 = 88.42 KJ

B. mass of water = 2Kg ( since density is 1 Kg/L ), S(water) = 4186 J/Kg-K, ΔT= 60K

Q( heat transferred to water ) = 2 * 4186 * 60 = 502.3 KJ

C. Assuming no heat loss :

Total heat energy transferred = 502.3 + 88.42 = 590.72 KJ ; Total time taken = 40mins = 2400 seconds

Power of stove = Q(total) / time taken = 246.13 Watts