An electric teapot has a heating element that operates at 120 v and with a current of 2.00 a. assuming the water absorbs all the energy delivered to the heating element, calculate the time interval (in s) during which the temperature of 0.891 kg of water rises from room temperature (23.0°c) to the boiling point. (the specific heat f water is 4,186 j/(kg°

1196.62 sec

Explanation:

V = electric potential difference at which teapot operates = 120 volts

i = current = 2.00 A

t = time of operation

m = mass of water = 0.891 kg

T₀ = initial temperature = 23.0 °C

T = final temperature = 100 °C

c = specific heat of water = 4186 J/(Kg °C)

Using conservation of energy

V i t = m c (T - T₀)

(120) (2.00) t = (0.891) (4186) (100 - 23.0)

t = 1196.62 sec

when it gains or loses electrons

the assestment will probley apele with cause and effect