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Rzqust [24]
3 years ago
15

A closed system contains 30g of gas. How much heat(in joules) is added to or rejected by the system to produce 5000 N-m of work

while the stored energy decreases by 200 J/g?
Physics
1 answer:
finlep [7]3 years ago
8 0

Answer : The heat rejected by the system is 1000 J

Explanation :

As per first law of thermodynamic,

q=\Delta U+w

where,

\Delta U = internal energy  of the system

q = heat  added or rejected by the system

w = work done of the system

First we have to determine the internal energy for 30 grams of gas.

As, 1 gram of gas has internal energy = 200 J

So, 30 grams of gas has internal energy = 200 × 30 = 6000 J

Now we have to determine the heat of the system.

q=\Delta U+w

\Delta  = -6000 J

w = 5000 N.m = 5000 J

Now put all the given values in the above formula, we get:

q=-6000J+5000J

q=-1000J

The negative sign indicate that the heat rejected by the system.

Hence, the heat rejected by the system is 1000 J

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m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).

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=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
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3 years ago
1. The current in a wire is 0.72A. Calculate the charge that passes through the wire in; a. 4 s b. 60 s c. 180 s d. 7 s e. 0.5 s
Vlad [161]
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t

Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C

Case II
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When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C

Case III
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When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C

Case IV
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When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C

Case V
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When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
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A would be the answer 
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Explanation:

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