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## Hot answers tagged thermodynamics

22

You blow away the flame from its fuel source. If you would blow less hard the flame might burn harder because more air is supplied to the flame (similar to a Bunsen burner). Because normally the flame of a candle gets its oxygen through a conventional airflow generated by the heat of the frame. The reason why the flame is blown away from the candle is ...

13

Combustion is a gas phase reaction. There are two requirements to generate a stable flame. Firstly the temperature must be high enough to vapourise the combustible material (wax in this case), and secondly the temperature must be high enough to generate the activation energy needed for the reaction. Heat is needed because gas phase molecules of wax and ...

4

If you look closely at the candle flame, you will notice that the flame hovers just over the wick, but does not touch it. This is because the flame is boiling the wax, which becomes a vapor, which then burns. All of these processes are driven by the heat from the flame. As you blow on the flame, you moving it away from the wax and disrupt this process. ...

3

If you rearrange your equation to solve for $Y$, $$Y=(X+125)\frac{-30+70}{360+125}-70$$ This reduces to $$Y=\frac{8X-5790}{97}=0.08247X+59.6907\approx0.08X-60$$ which is about what the textbook obtains. The difference in values you are getting is completely due to the approximation that the solution manual uses. If you use the exact values, as you did, ...

1

Heat source (or heat reservoir) and heat sink are terms used to describe thermodynamic cycles. It is easy to recognize them when You think about closed cycles. But vehicle engines are usually internal combustion type engines and as such perform open cycles. "Internal combustion engines are primarily heat engines" and "heat engines are often confused with ...

1

(Full disclosure, I didn't RTFA, and I don't have time to.) Just to review, $f\left(x,v,t\right)dxdv$ is the number of particles with positions between $x$ and $x+dx$ and velocities between $v$ and $v+dv$. First, why is there no integral over position? In principle, there should be. However, assuming that $d$ is small, \$f\left(x,v,t\right)\approx ...

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