Respuesta :
Answer:
1353.38 Watt
Explanation:
T₁ = Initial temperature of the house = 35°C
T₂ = Final temperature of the house = 20°C
Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s
m = mass of air in the house = 800 kg
Cv = Specific heat at constant volume = 0.72 kJ/kgK
Cp = Specific heat at constant pressure = 1.0 kJ/kgK
Heat removed
q = mCvΔT
⇒q = 800×720×(35-20)
⇒q = 8640000 J
Average rate of hear removal
[tex]Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W[/tex]
[tex]COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W[/tex]
∴ Power drawn by the air conditioner is 1353.38 Watt
Answer:
W_in = 1.353 KW
Explanation:
Given:
- Initial temperature of the house T_1 = 35°C
- Final temperature of the house T_2 = 20°C
- Time taken to cool the house dt = 38 min = 38×60 = 2280 s
- mass of air in the house m = 800 kg
- Specific heat at constant volume c_v = 0.72 kJ/kgK
- Specific heat at constant pressure c_p = 1.0 kJ/kgK
Find:
- Determine the power drawn by the air conditioner.
Solution:
- We will first compute the rate of heat removal from the room, we will use c_v due to a constant volume process, as follows:
Q_l = m*c_v*dT/dt
- In put values given:
Q_l = 800*0.72*(35-20) / 2280
Q_l = 3.7894 KW
- The relationship between the heat rejection and the COP of an air conditioner is given as:
COP = Q_l / W_in
W_in = Q_l / COP
W_in = 3.7894 / 2.8
W_in = 1.353 KW
- Hence the amount of power required for this process is 1.353 KW
