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trapecia [35]
4 years ago
8

A system absorbs 12 j of heat from the surroundings; meanwhile, 28 j of work is done by the system. what is the change of the in

ternal energy δeth of the system?'
Physics
2 answers:
lidiya [134]4 years ago
8 0
48 is ur answer bc i did this before
Shtirlitz [24]4 years ago
6 0

The change in the internal energy of the system after absorbing the energy is  \boxed{40\,{\text{J}}} .

Further Explanation:

Given:

The energy absorbed by the system from the surroundings is 12\,{\text{J}} .

The work done by the surroundings on the system is 28\,{\text{J}} .

Concept:

Since there is some amount of the work done on the system and the system also absorbs some amount of energy from the surroundings, the change in internal energy of the system is given by the First Law of Thermodynamics.

The First Law of Thermodynamics state that the total energy of an isolated system is always conserved. It means that the change in the internal energy of the system is equal to the sum of the amount of energy absorbed by the system and the amount of work done by the surroundings on the system.

\Delta U=Q+W

Here, \Delta U  is the change in internal energy of the system, Q  is the energy absorbed by the system and W  is the work done on the system.

Substitute the value of the energy absorbed and the work done in above expression.

\begin{aligned}\Delta U&=12\,{\text{J}}+28\,{\text{J}}\\&={\text{40}}\,{\text{J}}\\\end{aligned}

Thus, the change in the internal energy of the system is \boxed{40\,{\text{J}}} .

Learn More:

1. One consequence of the third law of thermodynamics is that <u>brainly.com/question/3564634 </u>

2. According to Charles’s law, for a fixed quantity of gas at constant pressure <u>brainly.com/question/7316997 </u>

3. Calculate the average translational kinetic energy (sometimes just called average kinetic energy) <u>brainly.com/question/9078768 </u>

Answer Details:

Grade: College

Subject: Physics

Chapter: Heat and Thermodynamics

Keywords:

System absorbs, heat from surroundings, work done, energy absorbed, change in internal energy, first law of thermodynamics,

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5 0
2 years ago
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
4 years ago
A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec
horsena [70]
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J
5 0
3 years ago
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