A glass tumbler of mass 400 g at room temperature, say 20 degree Celsius, is immersed into a tub of water maintained at 80 degree Celsius. If the temperature of the tumbler reaches that of the tub, how much heat does the tumbler absorb?
(Given : Specific heat of glass = 0•840x1000J/kg degree Celsius)
Answers
The density of water =1000kh/m
3
The density of water vapour =
0.6Kg/m
3
100°C
Pressure =1atp
The temperature =100°C
Estimation of the ratio of the molecular volume to the total volume occupied by the water vapour under the above condition of temperature and pressure.
T
1
P
1
V
1
=
T
2
P
2
V
2
T
1
P
1
V
1
=
T
2
P
2
V
2
(according to Charle's and Boyle's law)
where P
1
and P
2
$ be the density of water and water vapour respectively.
Since pressure and temperature are same and V
1
and V
2
be the volume of the water and water vapour respectively.
P
1
V
1
=P
2
V
2
1000×V
1
=0.6×V
2
V
2
V
1
=
10×1000
6
=6:10000
=3:5000
Answer:
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