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kumpel [21]
2 years ago
12

Write a short note on convection of heatI will give brainlist ​

Physics
1 answer:
makvit [3.9K]2 years ago
7 0
Convection is the process of heat transfer in fluids by the actual motion of matter. It happens in liquids and gases. It may be natural or forced. It involves a bulk transfer of portions of the fluid.


Hope that helps
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A certain parallel-plate capacitor is filled with a dielectric for which κ = 6.56. The area of each plate is 0.0830 m2, and the
mafiozo [28]

Answer:

4.86\times 10^{-7}\ J

Explanation:

Given:

Dielectric of the medium between the plates (k) = 6.56

Area of eac plate (A) = 0.0830 m²

Separation between the plates (d) = 1.95 mm = 0.00195 m [1 mm = 0.001 m]

Maximum electric field (E_{max}) = 202 kN/C = 202000 N/C [1 kN = 1000 N]

Permittivity of space (ε₀) = 8.854 × 10⁻¹² F/m

The maximum potential difference across the plates of the capacitor is given as:

V_{max}=E_{max}d\\\\V=(202000\ N/C)(0.00195\ m)\\\\V=393.9\ V

Now, capacitance of the capacitor is given as:

C=\dfrac{k\epsilon_0A}{d}\\\\C=\frac{6.56\times 8.854\times 10^{-12}\ F/m\times 0.0830\ m^2}{0.00195\ m}\\\\C=2.47\times 10^{-9}\ F

The maximum energy stored in the capacitor is given as:

U_{max}=\frac{1}{2}CV_{max}^2\\\\U_{max}=\frac{1}{2}\times (2.47\times 10^{-9}\ F)\times  393.9\ V\\\\U_{max}=4.86\times 10^{-7}\ J

Therefore, the maximum energy that can be stored in the capacitor is 4.86\times 10^{-7}\ J

6 0
2 years ago
A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.
ruslelena [56]

Answer:

C. W = 115.12\,Btu

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process (\eta), no unit, is:

\eta = \left(1-\frac{T_{L}}{T_{H}} \right) (1)

Where:

T_{L} - Temperature of the cold reservoir, measured in Rankine.

T_{H} - Temperature of the hot reservoir, measured in Rankine.

If we know that T_{H} = 1809.67\,R and T_{L} = 584.67\,R, then the energy efficiency of the ideal thermal process is:

\eta = 0.678

By First Law of Thermodynamics, we calculate the work output:

W = Q_{H}-Q_{L}

W = \frac{W}{\eta} -Q_{L} (By definition of efficiency)

Q_{L} = \frac{W}{\eta}-W

Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W(2)

Where:

Q_{H} - Heat received by the engine, measured in Btu.

Q_{L} - Heat rejected by the engine, measured in Btu.

W - Work output, measured in Btu.

If we know that \eta = 0.678 and Q_{L} = 55\,Btu, then the work output of the Carnot engine is:

W = \frac{Q_{L}}{\frac{1}{\eta}-1 }

W = 115.807\,Btu

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

5 0
3 years ago
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