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nataly862011 [7]
3 years ago
15

A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the

internal energy of the system?
Physics
1 answer:
Stels [109]3 years ago
3 0

Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

\Delta U=-12 J-(-28 J)=+16 J

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Answer:

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Explanation:

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now weight is given as

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Part b)

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Part c)

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8.918 = 1000(0.10^2 \times d) 9.8

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Answer:

Explained below

Explanation:

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Thermometry substance is mercury or alcohol

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Thermometry substance is Gas.

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Thermometry substance is Resistance wire.

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