Understanding Work and Energy Transfers in Gas Processes

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The discussion centers on calculating work done by a gas when its internal energy changes by 1500 J and heat transferred is -400 J. The initial assumption of work done being 1100 J is incorrect due to confusion over the signs of heat and work. The correct calculation shows that W(in) equals 1900 J, leading to the realization that W(out) must be considered as the opposite of W(in). The conversation highlights the importance of understanding energy conservation and the different conventions used in textbooks regarding work. Clarifying these concepts is essential for accurately determining energy transfers in gas processes.
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If the change in internal energy of a gas during a process is 1500 J and the heat transferred to the gas during the process -400 J, what is the work done by the gas?

I immediately though the answer to this question was 1100J, but it's not, I am completely confused by this. How can it not be 1100?
 
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Q(in) + W(in) = DeltaE

Q(in) is NEGATIVE , but DeltaE is POSITIVE.

so, what's W(in)?

Now notice that they're asking for W(out) ...
 
Q(in) + W(in) = DeltaE
-400 + W(in) = 1500
W(in) = 1900

So now how do I find W(out)? I figured that [W(in) + W(out)]/2 = change in internal energy which is how I came up with my original answer of 1100J, but that's incorrect.
 
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HINT: W(out) is the opposite to W(in)
 
Work transfers Energy OUT of one object and INto another ...
I was referring to Work that transfers Energy Into the gas as W(in).
Energy is conserved, so that W(in) would be W(out) for some other object.

Textbooks are notoriously flaky about which W they mean.
I envision all Energy transfers going into my sample, which raises its T.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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