A Carnot air conditioner takes energy from the thermal energy of a room at 61°F and transfers it as heat to the outdoors, which
1 answer:
Answer:
<em>14.07 units of joules is removed for each joules of electricity used</em>
Explanation:
The heated room temperature = 61 °F
The outside temperature = 98 °F
For conversion from fahrenheit to kelvin we use the equation
(32°F − 32) × 5/9 + 273.15
= 289.26
= 309.82
For air conditioning,
COP =
COP = = 14.07
This means that <em>14.07 units of joules is removed for each joules of electricity used</em>
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Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:
d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.
where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get
It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.