Uf a 2Kilowatt electric immersion heater is used to warm 200kg of water of 20°, Calculate a) the quantity of heat energy given out by immersion heat is 2 hours time. b) final temperature of water.​

(a) the quantity of heat energy given out is 1.44 x 10⁷ J

(b) the final temperature of the water is 37.14 ⁰C

Explanation:

Given;

power supplied to the heater, P = 2 kW = 2,000 W

mass of water, m = 200 kg

duration of the power supply, t = 2 hours = 2 x 3600s = 7,200 s

initial temperature of the water, t₁ = 20° C

specific heat capacity of water, c = 4200 J/kgC

(a) the quantity of heat energy given out by immersion heat is 2 hours;

Q = P x t

Q = 2,000 W  x 7,200 s

Q = 1.44 x 10⁷ J

(b) the final temperature of water

Q = mcΔt

Δt = Q / mc

Δt = (1.44 x 10⁷) / (200 x 4200)

Δt = 17.14

Δt = t₂ - t₁

Δt + t₁ = t₂

17.14 + 20 = t₂

37.14 ⁰C = t₂

i believe the answer is electric potential.

explanation: