I think this is AWESOME, but I think the last sentence of your conclusion is a bit off. <span> "If someone has an allergy to oil then they can still eat cake because applesauce makes an amazing substitute for oil." I think that you should say "This recipe is great for those who cannot eat/drink oil, the applesauce is an amazing substitute for oil."
I hope I helped! -Wajiha</span>
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁ ![ln[\frac{P_{1} }{P_{2} }]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7BP_%7B1%7D%20%7D%7BP_%7B2%7D%20%7D%5D)
= <em>mRT</em>₁ [∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039) ![ln[\frac{2.0}{1.0}]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7B2.0%7D%7B1.0%7D%5D)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> = 
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
It might be to late but the answer is C
