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$\bullet$$\bullet$ When a system is taken from state $a$ to state $b$ in Figure 15.34 along the path $a c b$ 90.0 $\mathrm{J}$ of heat flows into the system and 60.0 $\mathrm{J}$ of work is done by the system. (a) How much heat flows into the system along path adb if the tem along path adb if the work done by the system is 15.0 $\mathrm{J} ?$ (b) When the system is returned from $b$ to $a$ along the curved path, the absolute value of the work done by the system is 35.0 $\mathrm{J}$ . Does the system absorb or liberate heat? How much heat? (c) If $U_{a}=0$ and $U_{d}=8.0 \mathrm{J},$ find the heat absorbed in the processes $a d$ and $d b$ .

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Physics 101 Mechanics

Chapter 15

Thermal Properties of Matter

Temperature and Heat

The First Law of Thermodynamics

The Second Law of Thermodynamics

Rutgers, The State University of New Jersey

University of Washington

Hope College

Lectures

03:15

In physics, the second law of thermodynamics states that the total entropy of an isolated system can only increase over time. The total entropy of a system can never decrease, and the entropy of a system approaches a constant value as the temperature approaches zero.

03:25

The First Law of Thermodynamics is an expression of the principle of conservation of energy. The law states that the change in the internal energy of a closed system is equal to the amount of heat energy added to the system, minus the work done by the system on its surroundings. The total energy of a system can be subdivided and classified in various ways.

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On the part, a c b, the heat flows into the system is 90 joule, and the work done by the system is 60 joule. So we will find the change in internal energy between the point a and b. So in part, a c then beg to be amount of heat supplied into the system is 90. Joules. Work done by the system is 60 oleon. So we want to find the change in internal energy, as according to fossil athrodanamics we have come, is equal to w plus delta. U, as heat is added to the system, so we will take it as positive and work is done by the system. So work done. Will also be taken as positive, that is 60 joule, plus change in internal energy point, so we will get change in internal energy in part. A c b is equal to 30 joule now in part, a d b in part, a c b and a d b. We have same initial point and final point, as the change in internal energy does not depend upon pat, it only depends upon initial and final positions, so change in internal energy in part, a d b will also be equalled to change in internal energy of path. A c b that is 30 joules it so, in the part, a d b work done on the system is 15 joule and we want to calculate the amount of heat in part, a d v again by taking the first lot thermodynamics, we have q is equal To w plus delta: u here, w is 15 joule and change in internal energy is 30 joule, so we get c equal to 45 joule. So in the path 80 b, 4445 joule of energy is added to the system in the b part of this problem. It is given that the system returned from point b to point a in the curve path, with an absolute value of work done equal to 35 joule, and we want to calculate the amount of heat whether the heat absorb or liberated from the system. So the absolute value of work done is 35 joule from point b to the work is done in the system, as the system compressed so by taking fossil othermen makes that q is equal to w plus change in internal energy. Here, work done is in the system, so work done is negative. That is minus 35 joule plus is in the part i, the change in determined energy from point a to be was equal to 30 joule. So the change in turner from point b to a will be equal to minus 30 joule. So we have change in an energy equal to minus 30 joule. So we get c equal to minus 65 joule here negative sin shows that the system has liberated liberated 65 joules of energy when the system moves from point b to point a in the curve. Pathan cart! If we assume the internal energy of point, a is 0 and internal energy of point equal to a then we want to calculate the amount of her released are absorbed in the path, a d and in the path d again by taking first law thermodynami we have Is equal to w plus delta? U here delta? U that is internal energy changing, inter which is equal to internal energy. At point d, minus internal energy, a point a and work done from a to d, plus work done from d to b. That is equal to work done from a d b. It'S the work done from a to g plus work done from d to b. Is equal to 0 and there is no change in volume that is equal to work done from a d b that is equal to 15 joule, which is given in part b of this problem. So we get work done from a to d is equal to 15 joule. Now, by putting this value over here, we get. Cum is equal to work done from a to the which is 15 jo and change in internally from a 2 d. Is internal minus internal point, which is equal to 15 joule plus internal energy? At d? Is 8 joule and a point is 0 joule, so we get 15 joule plus an joule, so we get q equal to 23 joule, the stammato heat required in the path. A d here positive sign shows that heat is added to the system. Now in the path again by taking first lap throm, the anemic we have, from d to b is equal to work done from d to b, plus change in internal energy, from d to b. A volume is constant in the path d, so work done will be equal to 0, so heat in the path is equal to change in internal energy, from d to v, which is equal to internal energy point b, minus internal energy point d: if we take internal Energy at bits equal to 0, then intern energy. It will be equal to 30 joule, as in part of this problem we have delta. U, between point a b is equal to 30 point, so internal energy of point b is equal to 30 joule minus internal energy of point d is 8 joule. So by substracting these 2 values we get heat from pit d to b, is equal to 22 joule. So again, the heat is in positive. It shows that, from point d to point b, heat is added to the system.

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