The structure of a house is such that it loses heat ata rate of 3800 kJ/h per8C difference between the indoors andoutdoors. A heat pump that requires a power input of 4 kWis used to maintain this house at 248C. Determine the lowestoutdoor temperature for which the heat pump can meet theheating requirements of this house.

The lowest outdoor temperature for which the heat pump can meet the requirement of the house is -9.6°C

Explanation:

The coefficient of performance of a heat pump, COP, is given by the relation;

The rate of heat loss = 3800 kJ/h = 3800 ×1/60 × 1/60 kJ/s = 1.06 kJ/s

The power input = 4 kW = 4 kJ/s

The temperature indoors = 24 + 273.15 = 297.15 K

Therefore, we have;

4×297.15= (297.15- T₁) × 1.06×(297.15- T₁)

Which gives;

18/19·T₁²-627.32·T₁+92014.97=0

(T₁ -263.59)·(T₁ - 330.71) = 0

The solutions are;

T₁ = 263.59 K or T₁ = 330.71 K

The lowest temperature = 263.59 K which is -9.6°C

The lowest temperature = -9.6°C.

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )