A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the temperature in the house is 7°C when the heat pump is turned on, how long will it take for the heat pump to raise the temperature of the house to 22°C? Assume the entire mass within the house (air, furniture, etc.) is equivalent to 1500 kg of air. The constant volume specific heat of air at room temperature is cv = 0.718 kJ/kg·°C.

1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

it about 600 mph from what i calculated