Question

An electrical heating coil is immersed in 4.6kg of water 22°C.The coil, which has a resistance of 250Ω, warms the water to 32°C in 15min. What is the potential difference at which the coil operates?

Answer 1

Answer:

242.4 V

Explanation:

Applying,

Q = cm(t₂-t₁)................ Equation 1

Where Q = amount of  heat, c = specific heat capacity, m = mass, t₁ = initial temperature, t₂ = final temperature.

But this heat is produced by electrical power

Therefore,

P = Q/t

Where P = Electrical power, t = time

Also,

P = V²/R

Where V = voltage, R = resistance

Therefore

V²/R = Q/t......................... Equation 2

Substitute equation 1 into equation 2

V²/R = cm(t₂-t₁)/t................. Equation 3

Given: R = 250 ohms, m = 4.6 kg, t₁ = 22°C, t₂ = 32°C, t = 15 min. = (15×60) = 900 seconds

Constant: c = 4600 J/kg.°C

Substitute these values into equation 2

V²/250 = 4600(4.6)(32-22)/900

V²/250 = 235.11

V² = 250×235.11

V² = 58777.78

V = √58777.78

V = 242.4 V