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Thermal Physics

Cool goIITian

Joined:	30 Apr 2007
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18 May 2007 10:06:38 IST

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672

killer questions--------2

None

first qn. easy one

1. a metal block of density 6000kg/m3 and mass 1.2 kg is suspended through a spring of spring constant 200N/m . the spring-block system is dipped in a water kept in a vessel.the water has a mass of 260 g & the block is at a heigt 40 cm above the bottom of the vessel if the support to the spring is broken,what willl be the rise in the temperature of water.s for block is 250 J/Kg-K . heat capacities of the vessel & the spring are negligible....

2. a vessel of volume Vo contains an ideal gas at pressure po (o on foot) and temperature T . gas is continuosly pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant..the pressure of the gas being taken out equaks the pressure inside the vessel. Find

a)the pressure of the gas as a function of time..

b)the time taken before half of the original gas is pumped out..

3. a unioform tube closed at one ned.contains a pallet of mercury 10 cm long .when the tube is kept vertically with the closed end upward,the lenthg of the air column trapped is 20 cm.find the lenght of the air column trapped when the tube is inverted so that the closed end goes down....(its an easy one)

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Comments (3)

Ramandeep Singh

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Joined: 1 Jan 2007

Posts: 501

18 May 2007 16:23:32 IST

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hey i think i got ur 3rd one
look length of mercury column = 10cm = 100mm

in first case when tube is inverted pressure inside the tube = 760 - 100 = 660mm of mercury
in 2nd case when it is normal, pressure inside tube = 760 + 100 = 860mm of mercury

as pressure inversely proportional to volume

so P₁/P₂ = V₂/V₁
P₁/P₂ = l₂A / l₁A     {where l₁,l₂ r lengths of air columns and A is cross sectional
                                                                                                  area of tube}

l₂ = l₁ P₁/P₂
   = (20 660/860) = 15.34cm

pls rate if crrect

joy francis

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Joined: 19 Feb 2007

Posts: 1802

18 May 2007 18:55:30 IST

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in the first question the block is a cube right?

CyBorG

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Joined: 6 Jan 2007

Posts: 706

1 Jun 2007 15:29:12 IST

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Q1)The concept to be applied is the initial potential energy of the spring and the gravitational potential energy of the block gets converted into heat and the water and the block absorbs that heat.

Since the block is suspended completely in water we should calculate its apparent weight.

Volume of the block V=1.2/6000=0.0002m³

Weight of the water of same volume=0.0002*1000*10=2N

So apparent weight of the block is 12-2=10N

To calculate potential energy of the spring:

We know mg=kx (Since the spring gets extended due its APPARENT weight)

ie 10=200x

So x=0.05m

So potential energy of spring E₁=kx²/2=0.25J

Gravitional potential energy E₂=mgh=10*0.4=4J

So total initial energy=4.25J

This energy is converted into heat and it is absorbed by water as well as the block.

So using Q=mC

we get

4.25=1.2*250*

+0.26*4200*

=0.003⁰C

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