So change in internal energy $\Delta U=Q+W$; so in lifting an object you do work on it, thus increasing its internal energy. Is this correct?

Sorry if this is a stupid question.

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    $\begingroup$ I'd say that the internal energy remains unchanged and that there is no work done on the system. $\endgroup$ – thermomagnetic condensed boson Dec 22 '15 at 12:24
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    $\begingroup$ The added energy of lifting an object is not internal to the object, it is "between" or "among" the Earth and the object. Specifically, for a closed system the exact location of that system's energy can be thought of as at the system's center of mass. The internal energy of the object remains unchanged. $\endgroup$ – Digiproc Dec 22 '15 at 12:54

Not a silly question! Although your formula is the way that we often memorize the First Law, the true First Law is actually $$ \Delta (U + KE + PE) = Q + W_\text{non-grav} $$ where $KE$ and $PE$ are, respectively, the kinetic and potential energy and $W_\text{non-grav}$ is the work done by all forces except gravity. When elevating an object, work is done on the object, but its potential energy changes by the same amount (i.e., $\Delta U$ is indeed $0$). The First Law then breaks down to $$ \Delta PE = W_\text{non-grav} $$ ...which we know to be true from physics. The reason that the simplified form of the first law is so commonly referenced is that we often encounter systems in which $\Delta PE = \Delta KE = 0$

Edit: there is actually more than one way to look at this

  1. View Gravity as a force. In this view, the first law is still written as $$ \Delta (U +KE) = Q + W_\text{incl. grav} $$ but $W_\text{incl. grav}$ includes both the work done by the external force which elevates the object and by the force of gravity. These effects cancel out to zero.

  2. View gravity as a potential - i.e., track its effect by using potential energy rather than by considering it as a force which does work. In this view, the first law is as I wrote it above (first equation) and $W_\text{non-grav}$ the work done by all forces except gravity.

The two perspectives are equivalent because gravitational potential energy is defined by $\Delta PE_\text{grav} = - W_\text{grav}$.


Why is work done by a force when a body is lifted?

To counteract the gravitational force which would work had there not been the force & thus failing to move the object upward.

Does it increase the internal energy of the body?

Gravitational potential energy is a property of two objects; not of a single one. You can't distribute the potential energy as part of it as this object's & the remaining part to the other object. Potential energy is of the system and not of a single body constituting the system.

So, when you lift the body, you are increasing the potential energy of the system (earth-object) & not increase the internal energy of the object.


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