How Is Heat Calculated from Entropy and Temperature Changes in Thermodynamics?

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To calculate heat transferred to an ideal monatomic gas from entropy and temperature changes, the relationship dS = dQ/T can be used, leading to dQ = T dS. The slope of the temperature versus entropy graph provides T as a function of S, allowing for the integration of dQ. The user initially struggled with the integration process but eventually arrived at the correct answer after re-evaluating their approach. Despite achieving the correct result, they questioned the complexity of the method and sought a simpler alternative. The discussion emphasizes the importance of understanding the relationship between entropy, temperature, and heat transfer in thermodynamics.
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I am given the number of moles of an ideal monotomic gas. I am also given a chart with a graph of temperature plotted as a function of entropy.

One of the questions is: find the heat that was transferred to the gas.

How do I find that? I thought about saying:

dS = dQ/T <=> dQ = T dS

then I found the slope of the graph with respect to S and found the y-intercept. So now I had T in terms of S and I said:

dQ = (mS + b) dS

taking the integral of both sides I found Q and was able to easily integrate from initial entropy to final entropy. But I got the answer wrong. I don't want to do it all over again, unless I am sure this is the right approach. Is there an easier way?

This is problem #8 in Chapter 21 of Halliday/resnick/walker 6th edition.
 
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O.k. I overcame my sloth and did it again. I got the right answer!

But it seems overly involved. Is there an easier way?
 
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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