Question

A 2175 W oven is hooked to a 240 V source. (a) What is the resistance of the oven? in ohms

(b) How long will it take to boil 110 mL of water (initially at 20°C) assuming 80 percent efficiency? -- s

(c) How much will this cost at 10 cents/kWh? -- cents

Answer 1

here,

power input , Pi = 2175 W

potential difference , V = 240 V

a)

the resistance of own , R = V^2 / Pi

R = 240^2 /2175 ohm = 26.5 ohm

b)

mass of water , m = 110 g = 0.11 kg

let the time taken be t

using conservation of energy

Pi * e * t = m * Cw * ( 100 - 20) + m * Lv

2175 * 0.8 * t = 0.11 * ( 4186 * 80 + 2250000)

solving for t

t = 163.4 s

c)

the energy , E = P * t

E = 2175 * 163.4 /( 1000 * 3600) KWH

E = 0.0987 KWh

the cost , C = E * 10 cents

C = 0.987 cents