Help in understanding first way of changing the state of system

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?

• We never assume $T_A=0$. Indeed, that would make no sense. – Ivan Neretin Dec 21 '20 at 6:16
• 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. – theorist Dec 21 '20 at 6:47
• 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 $W_\mathrm{A}$ ?? – Poutnik Dec 21 '20 at 7:01
• @Poutnik $W_A$ is mechanical work . From 1st law of thermodynamics. U = q + w – srijan Sri Dec 21 '20 at 9:42
• @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 ? – Poutnik Dec 21 '20 at 10:07