Answer:
E=1824.81 V/m
Explanation:
Given that
Voltage difference = 5 V
Distance ,D= 3 mm
θ = 24°
As we know that electric filed given as
Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.
d= D cosθ
d= 3 cos24°
d = 2.74 mm
So
E=1824.81 V/m
Complete Question
A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).
J
(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)
MW
Answer:
The heat transferred is
The power is
Explanation:
From the question we are told that
Mass of the water per second is
The initial temperature of the water is
The boiling point of water is
The final temperature
The latent heat of vapourization of water is
The specific heat of water
The specific heat of stem is
Generally the heat needed each second is mathematically represented as
Then substituting the value
The power required is mathematically represented as
From the question
So