First Law of Thermodynamics

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a week ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

From first law of thermodynamics

? W = ? Q - ? U

? Q = n C p ? T

? U = n C v ? T

? W = n ( C p - C v ) ? T

? Q ? W = γ γ - 1 C p C p = γ

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

From first law of thermodynamics

? W = ? Q - ? U

? Q = n C p ? T

? U = n C v ? T

? W = n ( C p - C v ) ? T

? Q ? W = γ γ - 1 C p C p = γ

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

Δ Q = Δ U + Δ W

Q = Δ U + Q 5 Δ U = 4 Q 5 = n C v Δ T 4 Q 5 = 5 R 2 Δ T Δ T = 8 Q 2 5 R

Q = n c Δ T = 1 * C * 8 0 2 5 R C = 2 5 R 8 x = 2 5

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New answer posted

3 months ago

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alok kumar singh

Contributor-Level 10

Given                                                                           

QAB = +40J

QBC = 0J

QCA = -60J

WCA =?

WBC = -50J (on the gas)

UA = 1560J

1st Law on path A to B

WAB +   Δ U = Q

Þ 0 +    [ U B U A ] = 4 0

Þ UB = UA + 40

= 1560 + 40

UB = 1600 J

 1st Low on path B

...more

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3 months ago

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A
alok kumar singh

Contributor-Level 10

Sol. ? v

mv=16mv1

V1=V16

Δk loss =12mv2-12 (16m)V162

12mv21516

% loss =1516*100=93.75%

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