At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the
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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
To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,
Here,
I = Current
L = Length
B = Magnetic Field
= Angle between Force and Magnetic Field
But
Rearranging to find the Magnetic Field,
Here the force per unit length,
Replacing with our values,
Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T