Recent content by santimirandarp

I do. But isn't usually assumed that work in the environment is zero, and temperature is constant? And also, do you mean always ##Q_{env}=-Q_{sys}## and ##W_{sys}=-W_{env}##?

Yes, I am, but still miss something -I'll try to better understand what it is- that makes difficult to see that energy released as heat is absorbed as heat in a system. For example, why isn't this heat diminished by causing some orderly motion in the system? And also, why isn't it possibly that...

Thanks, yes but there is something else. Suppose we say '30J are transferred from the system to the environment as heat' it is completely abstract to me. What do we mean by energy here? It's not something easy to picture as work is, it seems. What do we mean by 'energy transferred as heat' that...

I couldn't derive it from 1st principle.

In an isothermal process, for an expanding gas ##\Delta U_{sys}=0## and ##Q=-W## but then, How can we evaluate ##Q_{surr} ##? It should be ##Q_{surr}=-Q_{sys}##, but I don't know how to show it in equations. If I try to get the result through the principles: ##\Delta U_{sys}=-\Delta U...

I'll ask the second part on a different question.

I've found it quoted as a general truth. Thanks.

It is not. The first law implies that universe internal energy is conserved and also: ##\Delta U_{sys}=(W+Q)_{sys}=-\Delta U_{surr}=-(W+Q)_{surr} ## so ##(W+Q)_{sys}=-(W+Q)_{surr} ## I don't see how it follows that dq=-dq_{surr}

It is not. The equality I showed is exactly what I don't understand ('why ...'). So I'm asking: where does the equality comes from? (and heat is not so simple as apples).

I know maths and conservation. But why do you say heat is conserved?

I'm sorry but I don't understand that answer

How do you derive that expression from the conservation of energy?

The question is: why ##dq_{surr}=-dq_{sys}##? q=heat, surr=surroundings, sys=system. Is there any simple way to understand this?

Adding yl=H/2 it'd be complete isn't it?

Good points. 1) I think is correct because t=L/v, time of flight is independent of the dimension. 2) The time intervals of acceleration are identical, that follows from v(t). 3) The integration was wrong y(t)= yl -amax*tl*(t-tl)+ amax*(t-tl)2/2 I'm not sure of this last step. So tired of...