Heat energy absorbed by a system in going through a cyclic process shown in figure is :-

Question

Heat energy absorbed by a system in going through a cyclic process shown in figure is :- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-s9pzepodktzrteek.jpg" /> (A) 107π J (B) 104π J (C) 102π J (D) 10- 3π J

Tardigrade Admin · Accepted Answer

In a cyclic, Δ U=0 From FLOT,Δ Q=Δ U+Δ W=0+Δ W= Area of closed curve ⇒ Q=π r2=π ((20/2))2kPa× litre =100π × 103× 10- 3J=100π J

Q. Heat energy absorbed by a system in going through a cyclic process shown in figure is :-

$1 0^{7} π J$

$1 0^{4} π J$

$1 0^{2} π J$

$1 0^{- 3} π J$

In a cyclic, $Δ U = 0$
From $F L OT, Δ Q = Δ U + Δ W = 0 + Δ W =$ Area
of closed curve
$\Rightarrow Q = π r^{2} = π (\frac{20}{2})^{2} k P_{a} \times$ litre
$= 100 π \times 1 0^{3} \times 1 0^{- 3} J = 100 π J$

Q. Heat energy absorbed by a system in going through a cyclic process shown in figure is :-

107πJ

104πJ

102πJ

10−3πJ

In a cyclic, ΔU=0 From FLOT,ΔQ=ΔU+ΔW=0+ΔW= Area of closed curve ⇒Q=πr2=π(220​)2kPa​× litre =100π×103×10−3J=100πJ

$1 0^{7} π J$

$1 0^{4} π J$

$1 0^{2} π J$

$1 0^{- 3} π J$

In a cyclic, $Δ U = 0$
From $F L OT, Δ Q = Δ U + Δ W = 0 + Δ W =$ Area
of closed curve
$\Rightarrow Q = π r^{2} = π (\frac{20}{2})^{2} k P_{a} \times$ litre
$= 100 π \times 1 0^{3} \times 1 0^{- 3} J = 100 π J$