A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejec
1 answer:
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Answer:
Option A
Solution:
As per the question:
Initially, voltage is
Current is 'I' A
Length of the wire is L
Now,
We know that:
(1)
where
A = Cross sectional area of the wire
From eqn (1), if other things are taken to be constant, then
R ∝ L (2)
Thus
When the wire is cut into two reducing the length to
Resistance, R' =
Now, when these wires are connected as described, the connection is in parallel, therefore, the equivalent resistance of the two wires:
Now, from Ohm's law:
Since, according to the question voltage is constant, thus
I ∝
I ∝
Thus
I becomes 4I
Answer:
a) The rate at which the cube emits radiation energy is 704.48 W
b) The spectral blackbody emissive power is 194.27 W/m²μm
Explanation:
Given data:
a = side of the cube = 0.2 m
T = temperature = 477°C
Wavelength = 4 µm
a) The surface area is:
According Stefan-Boltzman law, the rate of emission is:
b) Using Plank´s distribution law to get the spectral blackbody emissive power.