Electrical Energy and Power?

Question

The heating element of a coffeemaker operates at 120 V and carries a current of 2.00 A. Assuming that the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of water from room temperature (23.0 C) to the boiling point.

Answer 1

Energy delivered by the coffeemaker = 120*2*T = 240T joules

where

T = time

and since this is totally absorbed by water,

240T = MCp(delta T)

where

M = mass of water = 0.5 kg.

Cp = specific heat of water = 4186

delta T = 100 - 23 = 77

Substituting values,

240T = 0.5 * 4186 * 77

and solving for T

T = 0.5 * 4186 * 77/240

T = 671.50 sec = 11.19 minutes

Answer 2

The amount of energy needed to raise 0.500 kg of water from room temperature to the boiling point is:

Q = mc∆T = 0.500 kg x 4,186 J/kg°C x (100°C - 23°C) = 1.61 x 10^5 J

The power delivered is:

P = IV = 2.00 A x 120 V = 240 W

The time needed is:

t = Q / P = (1.61 x 10^5 J) / (240 W) = 671 s = 11.2 min

Answer 3

it will take 673.75 seconds........