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denis23 [38]
3 years ago
14

What volume of water at O°C can a freezer make into ice cubes in 1.0 h, if the coefficient of performance of the cooling unit is

6.0 and the power input is 1.7 kilowatt? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

volume of water is 0.1092 m3

Explanation:

We know that heat energy required to convert water into ice is given as

Q =mL

where L is Latent heat of fusion of ice is 336000 Jkg-1

Mass = \rho *V

           =1000*V

Q = 1000V*336*10^3

   Q = 336*10^6 V

ENERGY used by the freezer is given as

= 6.0*1.7*10^3 W = 10,200 w

energy provided per hour is Q = 10,200 W *1*60*60 s = 3.67*10^7 J

THEREFORE we have V

336*10^6 V = 3.67*10^7 j

v = .109 m3

volume of water is 0.1092 m3

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The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

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(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

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Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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