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10 Heat and Matter
W = 4πr
2
n − n
2/3
s
Also
W = 4π R
2 S
n
1/3
− 1
Then there will be a rise in temperature as energy is converted into heat.
Energy conservation gives
mc θ = 4π R
2 s
n
1/3
− 1
4π
3
R
3 cρ ρθ = 4π s R
3
1
r
−
1
R
∴ θ =
3s
ρc
1
r
−
1
R
10.65 W = 2 × 4π
r 2
2 − r 2
1
S
The factor of 2 arises as there are two surfaces.
W = 8π
(0.05)
2
− (0.01)
2
× 0.03
= 0.0018 J
10.66 The excess pressure must be equal to the pressure due to the water column
of depth h before the water leaks into the vessel.
2S
r
= ρ gh
∴ r =
2s
ρgh
=
2 × 0.073
1000 × 9.8 × 0.45
= 0.033 × 10
−3 m = 0.033 mm
10.67 Balancing the excess pressure in the bubble with the pressure due to a water
column of depth h
2S
r
= ρgh
∴ h =
2s
rρg
=
2 × 0.072
0.3 × 10 −3 × 1000 × 9.8
= 0.049 m = 4.9 cm
10.68 h =
2s
rρg
=
2 × 0.073
0.2 × 10 −3 × 1000 × 9.8
= 0.07449 m = 7.45 cm
The tube is inadequate as it is only 6 cm long. Water will not overflow. But the
radius of meniscus r 1 would now increase such that the following condition
is satisfied: