Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
In the diagram shown Q text iaf =80 cal and W text iaf =50 cal. If W=-30 cal for the curved path fi, value of Q for path fi, will be
Q. In the diagram shown
Q
iaf
=
80
cal and
W
iaf
=
50
cal. If
W
=
−
30
cal for the curved path
f
i
, value of
Q
for path
f
i
, will be
4173
194
Thermodynamics
Report Error
A
60 cal
10%
B
30 cal
14%
C
-30 cal
21%
D
-60 cal
55%
Solution:
From process iaf
Find
Δ
U
first,
Δ
Q
=
Δ
W
+
Δ
U
80
=
50
+
Δ
U
30
c
a
l
=
Δ
U
Use this
Δ
U
for process if
Δ
Q
=
Δ
W
+
Δ
U
Δ
Q
=
−
30
+
(
−
30
)
=
−
60
c
a
l