Respuesta :
Answer:
The correct option is;
(c) 64W
Explanation:
Here we have the Coefficient Of Performance, COP given by
[tex]COP = \frac{Q_{cold}}{W} = 3.1[/tex]
The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have
[tex]Q_{cold}[/tex] = m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J
Therefore, plugging in the value for [tex]Q_{cold}[/tex] in the COP equation we get;
[tex]COP = \frac{197.39 }{W} = 3.1[/tex] which gives
[tex]W = \frac{197.39 }{3.1} = 63.674 \ J[/tex]
Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.
The power input to the refrigerator is approximately 64 W.
Answer:
Win = 64 W ... Option C
Explanation:
Given:-
- The water is cooled in the refrigerator with delta temperature, ΔT=(23 - 6 )
- The flow rate of the refrigerated water is flow ( m ) = 10 kg/h
- The COP of the refrigerator is = 3.1:
Find:-
the required power input to this refrigerator is
Solution:-
- The COP - The coefficient of performance of a refrigerator is a quantity that defines the efficiency of the system. The COP is given as:
COP = QL / Win
Where,
QL : The rate of heat loss
Win : The input power required
- The rate of heat loss can be determined from first law of thermodynamics.
Qin - Wout = flow (m)*c*ΔT
Where,
Qin = - QL ... Heat lost.
c : The heat capacity of water = 4,200 J / kg°C
- There is no work being done on the system so, Wout = 0
-QL = flow (m)*c*ΔT
-QL = ( 10 / 3600 )*4200*( 6 - 23 )
QL = 198.33 W
- The required power input ( Qin ) would be:
Win = QL / COP
Win = 198.33 / 3.1
= 63.97 W ≈ 64 W
