35 - Heat and Thermodynamics Questions Answers
IN THE PREVIOUS QUES P2V3 =K BUT SORRY I AM NOT GETTING HOW PV3/2 =K
take square root on both the sides
DURING AN ADIABATIC PROCESS IF THE PRESSURE OF AN IDEAL GAS IS PROPTIONAL TO THE CUBE OF ITS ABSOLUTE TEMP. THEN RATIO OF SPECIFIC HEATS AT CONSTANT PRESSURE & AT CONST. VOL IS?
ANS 3/2
P α T3
so P α (PV)3
so P2V3 = K
or PV3/2 = K
so on the basis of PVγ = K calculate γ
the density of two substances are in ratio 5:6 & specific heat are in ratio 3:5. their thermal capacity per unit volume will be in ratio?
ans 1:2
thermal capacity = ms = Vds
so thermal capacity per unit volume = ds
now calculate answer
it takes 6 hours for ice layer thickness to grow from 1cm to 2cm on a lake in Siberia, ideally the time taken for thickness to grow from 3cm to 4cm?
ans 14 hrs
for this type of question relation between time and thickness will be
[x2] α t
so 22 - 12 = k6
and 42 - 32 = kt
on dividing we will get 3/7 = 6/t
or t = 14
Q HEAT & WORK ARE EQUIVALENT. THIS MEANS
TEMP OF BODY CAN BE INCREASED BY DOING WORK ON IT. PLS TELL HOW THIS STAT. IS CORRECT.Q
Q IF H IS SUPPLIED TO IDEAL GAS IN ISO PROCESS. THEN W= +ve HOW? BECAUSE IF ISO THEN U= 0 & EQN IS Q=W+0 & H IS SUPPLIED SO Q=-veSO W =-ve.so ans shd be -veBUT ANS GIVEN IS +ve PLS TELL ANS WITH EXPL. THANKS FOR PRE SOLUTIONS .
This means heat can be used to make work and work done produces heat
By doing 4.2 joule work 1 cal. heat will be produced in a body so due to this 1 cal, temp. will increase
In physics heat is supplied means Q will be +ve so W will also be positive hence is correct. A system is placed in refrigrator for absorbing heat from the body, this process will have Q -ve.
Q IF Al DIPPED IN H2O AT 10*C THEN FORCE OF BUOYANCY WILL INCREASE OR DECREASE?
Q TEMP. OF A SOLID OBJECT IS OBSV. CONST .DURING A PERIOD . IN THIS PERIOD
a)HEAT MAY HAVE BEEN SUPPLIED TO IT
b) H MAY HAVE BEEN EXTRACTED FROM IT
c) no H SUPPLIED TO IT
d)NO H EXTRACTED FROM IT
more than one opt is correct . pls explain your each opt with expln.
It depends that either Al or water will change its density more rapidly, because Bouyancy is based on densities of solid and liquid
Answer : a, because it may be an isothermal process
b, because in isothermal process, surrounding can also give heat to the system.
c, may be the process is adiabatic
d, may be adiabatic.
Q gas of certain mass at 273K is expanded to 81 times its vol. under adia. condition. if gamma = 1.25 for gas then its final temp. = ?
Q for a const. vol. gas thermometer one should fill gas at high temp & low pressure WHY ?
Q iF TEMP of uniform rod is slightly increase by del T its moment of inertia about 90* bisector increase BY = ? (ANS 2alpha I del T ).
Q Al sphere is dipped into H2O at 100*C if temp is increased forse of buoyancy will inc. OR decreased WHY =?
DENSITY OF A SUB. IS 10g/cc at 0*C &at 100*C d= 9.7 g/cc . coeff. of linear expantion is =?
Answer : For adiabatic process TVγ-1 = K
so 273*V0.25 = T*(81V)0.25
so 273/3 = T
Answer : Because real gases perform ideal behaviour in this range.
Answer : Inertia about perpendicular bisector = 1/12 ML2
on increasing temperature by dT new inertia = 1/12 M[L(1+αdT)]2
so increment in inertia = 1/12 ML2 [(1+αdT)2-1-] = 1/12 ML2 [ (αdT)2 + 2αdT ] now neglect (αdT)2 because it is small term
Answer : Since temperature is already 100 degree centigrade so on increasing it further density of Al will increase rapidly in comparision to density of water so Buoyancy force will be increased.
Answer : use the formula d100 = d0 / (1+γ100) here γ=3α calculate α
Q Agraph having temp *C along xaxis& *F along y axis. if graph is straight then it intercept +ve y axis. WHY ?
(F-32)/9 = t/5
so F = 9t/5 + 32
on taking t = 0 we will get F = 32
thats why it makes an intercept of 32 in y axis
a hollow sphere of mass M (in K.G.) radius R (in m.) is rotating with angular frequency omega(in radsec) . it suddenly stops rotating and 75% of the K.E. converted into heat energy . if S joulekg-kelvin is the spesific heat of material of the sphere, the rise of temperature of the sphere is ?
Rotational K. E. = 1/2 I ω2
so 0.75 [1/2 I ω2] = M S dT
and I = 2/3 M R2
now solve
two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end connected to the same vessels. Let M1 and M2 g/s be the rate of melting of ice in the two cases respectively, the ratio M1/M2 is
(1) 1:2
(2) 2:1
(3) 4:1
(4) 1:4
in first case rods are taken in parellel so net length and area will be l and 2A with conductivity coefficient K
similarly in second case rods are taken in series so net length and area will be 2l and A with conductivity coefficient K
now use H = KA(θ1- θ2)/L and H = dQ/dt = mL/t L is latent heat which is fixed in both the cases