Answer:
(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
Step-by-step 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