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

d eras and periods

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft