State how heat loss by radiation is minimized in a thermos flask
1 answer:
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Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
Answer:
The system loses 90 kJ of heat
Explanation:
We can answer the question by using the 1st law of thermodynamics, which states that:
where
is the change in internal energy of the system
is the heat absorbed by the system (positive if absorbed, negative if released by the system)
is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)
In this problem, we have:
is the work done (negative, because it is done by the surrounding on the system)
is the increase in internal energy
Using the equation above, we can find Q, the heat absorbed/released by the system:
And the negative sign means that the system has lost this heat.