Given:
Temperature of water, 
= 
=273 +(-6) =267 K
Temperature surrounding refrigerator, 
= 
=273 + 21 =294 K
Specific heat given for water, 
= 4.19 KJ/kg/K
Specific heat given for ice, 
= 2.1 KJ/kg/K
Latent heat of fusion, 
= 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max 
= 
= 
= 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max 
= 
= 10.89
Answer:
Hello Monk7294!
Answer:
Employee education
Explanation:
The most important countermeasure for social engineering is employee education. All the employees should be trained to keep confidential data safe. As a part of security education, organizations have to provide timely orientation about their security policy to new employees. The security policy should address the consequences of the breaches.
<em>- I Hope this helps Have an awesome day!</em>
<em>~ Chloe marcus <3</em>
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is
put here value with
turbine is insulated so q = 0
so here
solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
Answer:
The current through the coil is 2.05 A
Explanation:
Given;
number of turns of the coil, N = 1
radius of the coil, r = 9.8 cm = 0.098 m
magnetic moment of the coil, P = 6.2 x 10⁻² A m²
The magnetic moment is given by;
P = IA
Where;
I is the current through the coil
A is area of the coil = πr² = π(0.098)² = 0.03018 m²
The current through the coil is given by;
I = P / A
I = (6.2 x 10⁻² ) / (0.03018)
I = 2.05 A
Therefore, the current through the coil is 2.05 A
