Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is

Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:

So we equate the COP of our heater with COP of Carnot heater

Rearrange the equation

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C
Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Answer and Explanation:
The explanation is attached below
E. Parts they don’t resemble
Answer:
2.5 is the required details