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There is a statement in my book:

THERMODYNAMICS

One way: We do some mechanical work, say $\pu{1 kJ},$ by rotating a set of small paddles and thereby churning water. Let the new state be called $\mathrm{В}$ state and its temperature, as $T_\mathrm{B}.$ It is found that $T_\mathrm{B} > T_\mathrm{A}$ and the change in temperature, $\Delta T = T_\mathrm{B} - T_\mathrm{A}.$ Let the internal energy of the system in state $\mathrm{B}$ be $U_\mathrm{B}$ and the change in internal energy, $\Delta U = U_\mathrm{B} - U_\mathrm{A}.$

By rotating a small set of peddles, let us say we do mechanical work of $\pu{1 kJ}.$ Let the new state then be $\mathrm{B}$ and $\mathrm{A}$ as initial state.

It is found that $T_\mathrm{B} > T_\mathrm{A}.$ Let internal energy of system be $U_\mathrm{B}.$

They say that $\Delta U = U_\mathrm{B} - U_\mathrm{A}.$ Do we assume $T_\mathrm{A} = 0?$

We know $\Delta U = q + W.$ So, is $q = T_\mathrm{B} - T_\mathrm{A}?$

Should we say that work, i.e $W_\mathrm{A} = 0$ and same for $W_\mathrm{B} = \pu{1 kJ}?$

What is the difference in saying $\Delta U$ as internal energy and amount of mechanical work?

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  • $\begingroup$ We never assume $T_A=0$. Indeed, that would make no sense. $\endgroup$ – Ivan Neretin Dec 21 '20 at 6:16
  • $\begingroup$ If you are not changing the composition, q (the heat flow) depends on ΔT, the heat capacity of the substance, and the amount of substance; it is not equal to ΔT. It would make no sense for it to be equal to ΔT, since ΔT has units of degrees, while q has units of energy. I could give you the formula, but the fact that you don't realize it doesn't make sense to equate q to ΔT tells me you haven't yet read the section of your textbook on this subject. Thus I think it would be more to your benefit if you first read it and then asked questions if something doesn't make sense. $\endgroup$ – theorist Dec 21 '20 at 6:47
  • $\begingroup$ A $T$ in context of thermodynamics always means absolute temperature, it cannot be T=0, as this state is reachable only after infinite number of steps. $q = C \cdot ( T_\mathrm{B} - T_\mathrm{A})$, where $C$ is a heat capacity of the system $C = \frac{đq}{dT}$. What is $W_\mathrm{A}$ ?? $\endgroup$ – Poutnik Dec 21 '20 at 7:01
  • $\begingroup$ @Poutnik $W_A$ is mechanical work . From 1st law of thermodynamics. U = q + w $\endgroup$ – srijan Sri Dec 21 '20 at 9:42
  • $\begingroup$ @user282657 I have not meant $W$ generally, but $W_\mathrm{A}$ ( and $W_\mathrm{B}$ ) particularly. Is not there 1 single value of $W$ to get us from the state A to the state B along a particular path ? $\endgroup$ – Poutnik Dec 21 '20 at 10:07

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