Friction and Heat

Classical Mechanics Level 1

A block sliding along a tabletop is brought to rest by friction. If the friction does 50 J of work on the block in the process and the internal energy of the block increases by 30 J, how much heat (in Joules) is released to the block's surroundings?

The answer is 20.

1 solution

July Thomas
Mar 30, 2016

By the first law of thermodynamics

$\Delta U=\Delta Q+\Delta W$

where $\Delta U$ is the internal energy of the system, $\Delta Q$ is the heat supplied to the system, and $\Delta W$ is the work done to the system.

Here, $\Delta W=50$ J and $\Delta U=30$ J.

30 J = $\Delta Q$ + 50 J

$\Delta Q$ = -20 J

But $\Delta Q$ is the heat added to the block, so the heat released to the surroundings is the opposite: +20 J.

But the equation is dq=du+dw

Mohammed Faheem - 4 years, 9 months ago

By the first law of thermodynamics, dQ= dU + dW we know, dU = the internal energy of the system dQ = the heat supplied to the system, And dW = the work done to the system. Here, dW = 50J , dU = 30J By law, dQ= 50J + 30J dQ = 80J

80J Heat is released to the block’s surroundings. [Solved]

Mehedi Hasan - 3 years, 7 months ago

No, the first law of thermodynamics is ΔU=ΔQ+ΔW where ΔU is the internal energy of the system, ΔQ is the heat supplied to the system, and ΔW is the work done to the system. And you wrote another version of the law, where ΔW is the work done by gas itself, so the work done to the system is negative, because neither we do this work, but the system do it. So we have this equation. ΔU=ΔQ+(-ΔW) ΔQ=ΔU+ΔW In our case the work is not done by system, but by us to the system, so the answer is 30 J.

Вячеслав Чаунин - 1 year, 4 months ago

80 wil be the answer beacuse first law is dq=du+dw source: fundamentals of physics

Tazdidul Islam - 9 months, 3 weeks ago

Friction and Heat

The answer is 20.

1 solution

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