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vlada-n [284]
3 years ago
7

How does rusting differ from a combustion reaction

Physics
1 answer:
olga nikolaevna [1]3 years ago
7 0
Cool good luck with this 73737383+8282828=swo
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Andrei [34K]

screwdriver :)

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If you take a bar magnet with both n & s poles and cut it in half, you get
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Which of the following are simple machines?<br> Select all that apply.
scoundrel [369]

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Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.659 kJ of heat. It shrinks on
bixtya [17]
<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

<em />

<em />

5 0
4 years ago
What is the change in internal energy (deltaE ) of a system when it loses 76.0 J of heat while the surroundings perform 29.0 J o
lisabon 2012 [21]

Answer:

-47 J

Explanation:

Given that,

Heat loss = -76 J (negative for loss)

Work done by the surroundings = -29 J

We need to find the change in internal energy of a system.

The first law of thermodynamics is given by :

\Delta U=Q-W

W is work done by the system

Putting all the values,

\Delta U=(-76)-(-29)\\\\=-47\ J

Hence, the change in internal energy is -47 J.

3 0
3 years ago
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