Aheat pump has a coefficient of performance that is 60% of the carnot heat pump coefficient of performance. the heat pump is used to heat a home to 24.0°c during the winter with the low temperature reservoir at the outdoor temperature. at which outdoor temperature would it be more efficient to add the energy directly to the interior of the home than use it to run the heat pump?

=118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump=

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP=

in the question

⇒

heat is added directly to be as efficient as via heat pump

and T_H= 24° C= 297 K

on calculating the above equation we get

=118.8 K

the outdoor temperature for efficient addition of heat to interior of home

=118.8 K= 154.2°C

there are no correct statementd on the list of choices you provided.

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia