First Law of Thermodynamics

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Consider a system composed of a gas+stirrer at state 1, Work is done on the system by a rotating shaft, the stirrer activates high speed generating a significant amount of heat. The system is undergoing a process because the set of measurable properties to define its state are changing, If I disconnect the source delivering work, and let the gas cool down by putting my system in a cooler surrounding, it will release heat. If the assumption: Work done of my system=heat lost to the surrounding is true then consequently the initial state and final states are the same, but I cant accept this assumption as truth. For heat transfer to take place I have to place my system in a cooler surrounding, how is the quantity of heat transferred equal the work I added to my system
 
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The initial system is at the same temperature as the surrounding environment? I'll assume so.
Work done of my system=heat lost to the surrounding is true then consequently the initial state and final states are the same
If you only look at gas+stirrer this is true.
For heat transfer to take place I have to place my system in a cooler surrounding
Cooler than the system after stirring.
how is the quantity of heat transferred equal the work I added to my system
How could it be different? Energy is conserved and your system has a fixed energy to temperature relation (assuming nothing else happens in your gas, no chemical reactions and so on). If initial and final temperature are the same then initial and final energy are the same, and energy out = energy in.
 
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At the end of the stirring process, the temperature of the gas is higher than its initial temperature: $$\Delta U_1=nC_v(T_1-T_0)=-W$$where -W is the work done by the stirrer on the gas. During the cooling step of your process, the gas cools down, and its final temperature is equal to its initial temperature (the temperature of the surroundings) prior to the stirring: $$\Delta U_2=nC_v(T_2-T_1)=nC_v(T_0-T_1)=Q=-\Delta U_1=W$$
When we say that the initial stirring "generates heat," what we are really very loosely saying is that the internal energy of the system increases, rather than any actual heat transfer being involved.
 

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