Answer:
The entropy change of the sample of water = 6.059 x 10³ J/K.mol
Explanation:
Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol
Mathematically, entropy is expressed as
ΔS = ΔH/T....................... Equation 1
Where ΔH = heat absorbed or evolved, T = absolute temperature.
<em>Given: If 1 mole of water = 0.0018 kg,</em>
<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>
<em>T = 100 °C = (100+273) K = 373 K.</em>
<em>Substituting these values into equation 1,</em>
<em>ΔS =2.26x 10⁶/373</em>
ΔS = 6.059 x 10³ J/K.mol
Therefore the entropy change of the sample of water = 6.059 x 10³ J/K.mol
C. electrical energy is transformed into light and heat energy
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
Answer:
KE = 2.535 x 10⁷ Joules
Explanation:
given,
angular speed of the fly wheel = 940 rad/s
mass of the cylinder = 630 Kg
radius = 1.35 m
KE of flywheel = ?
moment of inertia of the cylinder

=
= 574 kg m²
kinetic energy of the fly wheel

KE = 2.535 x 10⁷ Joules
the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules
when a hole is made at the bottom of the container then water will flow out of it
The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.
By Bernoulli's equation we can write

Now by equation of continuity


from above equation we can say that speed at the top layer is almost negligible.

now again by equation of continuity


here we have


now speed is given by

