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?

Short Answer

Expert verified

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

Step by step solution

01

Given data

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

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

02

Find an expression to calculate the heat transfer from a system.

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.

03

Calculate the heat transfer using the given data.

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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Most popular questions from this chapter

(a) On a winter day, a certain house loses 5.00×108 J of heat to the outside (about 500,000 Btu). What is the total change in entropy due to this heat transfer alone, assuming an average indoor temperature of 21.0 ºC and an average outdoor temperature of 5.00 ºC ? (b) This large change in entropy implies a large amount of energy has become unavailable to do work. Where do we find more energy when such energy is lost to us?

Explain how water’s entropy can decrease when it freezes without violating the second law of thermodynamics. Specifically, explain what happens to the entropy of its surroundings.

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