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nevsk [136]
3 years ago
9

Joe noted that the temperature in his aquarium dropped overnight. He concluded that thermal energy moved from the aquarium to th

e air around it. Does this validate or invalidate the Law of Conservation of energy?
Physics
2 answers:
alukav5142 [94]3 years ago
5 0

Answer:

 validate                                            

Explanation:

<u>Law of Conservation of energy</u>

According to the law of conservation of energy, energy of a body or a system gets transferred to another system as different energy. We cannot create energy nor we can destroy energy. Energy is always transformed into other forms of energy.

In the context, Joe noticed that the aquarium's temperature dropped during the night and he concluded that according to the law of conservation of energy, the energy of the aquarium in terms of thermal energy is being transferred to the surrounding air.

This is because the aquarium is considered to be an open system where energy can come in and move out.

Hence Joe's statement is valid.

Sedaia [141]3 years ago
5 0

Answer:

It validates the law of conservation of energy

Explanation:

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Answer:

10 m/s

Explanation:

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We have to make v the subject so we should rearrange the equation

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v²= \frac{K}{0.5*m} (use algebra)

v²= 7/0.5×0.140

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We can confirm this by using the kinetic energy formula.

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What is an Environmental Footprint?
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The sun is currently a ____________ star.
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Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, −10°C with a mass flow rate of 15 kg/min and is com
Juli2301 [7.4K]

Answer:

a) W=85.225 kW

b) 0.02\frac{kW}{K}

Explanation:

First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):

W+m*h_1=Q+m*h_2\\W=Q+m*(h_2-h_1)

Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, h_1=h(200kPa,-10C)=1440.6\frac{kJ}{kgK},  h_2=h(1200kPa,140C)=1757.5\frac{kJ}{kgK}

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W=6kW+15\frac{kg}{min}*\frac{min}{60s}*(1757.5-1440.6)\frac{kJ}{kg}\\W=85.225kW

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dS=\frac{dQ}{T}

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