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Q2PE

Expert-verifiedFound in: Page 550

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**How much heat transfer occurs from a system, if its internal energy decreased by \({\bf{150}}\;{\bf{J}}\)while it was doing 30.0 J of work?**

The amount of heat transferred from the system is \(180\;{\rm{J}}\).

Change in internal energy, \(\Delta U = - 150\;{\rm{J}}\)

Work done, \(W = - 30\;{\rm{J}}\)

**According to the first law of thermodynamics, the change in internal energy of a system is equal to the difference of heat transfer occurring in the system and the net work done. It is mathematically expressed as follows;**

\(\begin{aligned}{}\Delta U = Q - W\\Q = \Delta U + W\end{aligned}\)

Here,\(\Delta U\)is the change in internal energy, \(Q\)is the net heat transfer and \(W\) is the net work done.

System does the work, hence negetive. The value of the change in internal energy and the work done is given.

\(\begin{aligned}{}\Delta U &= - 150{\rm{ J}}\\W &= - 30{\rm{ J}}\end{aligned}\)

Substituting these values in the expression for heat transfer will give the amount of heat transferred from the system.

\(\begin{aligned}{}Q &= \Delta U + W\\ &= - 150{\rm{ J}} + ( - 30{\rm{ J }})\\ &= - 180{\rm{ J}}\end{aligned}\)

So, \(180\;{\rm{J}}\)of heat is transferred from the system.

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