Answer:
Given,
Temperature;
T = 393;;K
Convert to Celcius;
T = (393-273) degrees
T = 120°C
Using Table A-4 (Saturated water - Temperature table), at T = 120 C;
vf = 0.001060 m³/kg
vg = 0.89133 m³/kg
Quality is given as;
75% = 0.75
Specific volume is given as;
v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)
v= 0.66876 m³/kg
We know;
v = V/m
0.66876 = 100/m
m = 149.53 kg
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
Yes
Explanation:
Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.
We know that
COP of heat pump= 1 + COP of refrigeration
It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
