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rodikova [14]
3 years ago
7

In an experiment, 108 J of work was done on a closed system. During this phase of the experiment, 79 J of heat energy was added

to the system. What was the total change in the internal energy of the system?
Physics
1 answer:
amid [387]3 years ago
7 0

Answer:

187 J

Explanation:

First Law of Thermodynamics :

ΔQ = ΔW + ΔU

ΔQ : Heat. If it added to system then positive and if it is rejected by system then negative.

ΔW : Work. If it done by the system then positive and if it is done on system then negative.

ΔU : Internal Energy. If it positive then temperature of system increased and if it is negative then temperature of system decreased.

ΔQ = 79 J

ΔW = - 108 J

ΔU = ?

substituting the value in the equation:

79 = -108 + ΔU

∴ ΔU = 187 J

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Anna11 [10]

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a. 2143 turns/m

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Explanation:

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\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m

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3 years ago
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