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I'm confused about what exactly is Q and U and their signs.

Consider a block initially having some kinetic energy which we stop and we want to find by how much amount its temperature increases.

Then Q=dU+W where dU is change in internal energy. (I'm using the sign convention that heat absorbed by the system is positive and work done by the system is positive).

Now we do −ve work on the system while stopping it so the work done by the system is +ve. Is this part right?

Now I'm really confused as to what to do next.

I assumed system is not absorbing or giving out any heat so that dU=−W and since work is positive that means dU is negative and so this means temperature should very counter intuitively decrease.

Isn't dU for the block mc(T(f)−T(i)) where T(f) is final temperature and T(i) the initial temperature , m is mass and c is specific heat capacity or is that wrong?

A different approach I thought of was that dU=0 so Q=W and now we can let Q=mc(T(f)−T(i)) to find final temperature which'll come more than initial. This seems to work fine but I don't know why. Why should dU be 0 when temperature is increasing?

The first approach seems to work fine with a different example of a human climbing an incline. There, work done by the human is positive and heat is 0 (or negative in the case human sweats, again I'm confused how should i go about deciding Q) So dU is negative (and even more negative if you consider sweating , again as expected) which is true since fat is used up.

I'll provide one final example. Suppose a gas is in a container initially having some velocity. Then we stop it. We've to find the final temperature.

Here again W is positive , now I'm again confused what to assume about Q. If we assume no heat exchange , Q=0 and so we've du=−W so dU is a negative quantity. But dU of ideal gas is 3/2nR(T(f)−T(i)) so this suggests final temperature is actually less and this seems wrong.

I looked in different textbooks , but none has good discussion on this. Any help is appreciated.

Consider a block initially having some kinetic energy which we stop and we want to find by how much amount its temperature increases.

Then Q=dU+W where dU is change in internal energy. (I'm using the sign convention that heat absorbed by the system is positive and work done by the system is positive).

Now we do −ve work on the system while stopping it so the work done by the system is +ve. Is this part right?

Now I'm really confused as to what to do next.

I assumed system is not absorbing or giving out any heat so that dU=−W and since work is positive that means dU is negative and so this means temperature should very counter intuitively decrease.

Isn't dU for the block mc(T(f)−T(i)) where T(f) is final temperature and T(i) the initial temperature , m is mass and c is specific heat capacity or is that wrong?

A different approach I thought of was that dU=0 so Q=W and now we can let Q=mc(T(f)−T(i)) to find final temperature which'll come more than initial. This seems to work fine but I don't know why. Why should dU be 0 when temperature is increasing?

The first approach seems to work fine with a different example of a human climbing an incline. There, work done by the human is positive and heat is 0 (or negative in the case human sweats, again I'm confused how should i go about deciding Q) So dU is negative (and even more negative if you consider sweating , again as expected) which is true since fat is used up.

I'll provide one final example. Suppose a gas is in a container initially having some velocity. Then we stop it. We've to find the final temperature.

Here again W is positive , now I'm again confused what to assume about Q. If we assume no heat exchange , Q=0 and so we've du=−W so dU is a negative quantity. But dU of ideal gas is 3/2nR(T(f)−T(i)) so this suggests final temperature is actually less and this seems wrong.

I looked in different textbooks , but none has good discussion on this. Any help is appreciated.

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