i want hc verma solutions for calorimetry full not 3 questions
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Complete Solution of HC Verma Concept of Physics-II Chapter 25 Calorimetry Has Been Uploaded
A rod of length 1m rest on a smooth horizontal base.When heated the length becomes 1.01 then find the strain doveloped.
A rod of length 1metre rest on a smooth horizontal base.When rod is heated its length becomes 1.01 then find the strain doveloped.
no strain will be developed. think of strain as elasticity. there should be a restoring force present. here there aint any.
Solve this question:
An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and length l. Left part of the cylinder contains 1 mole of helium and right part contains 0.5 mole each of helium and oxygen. If initial pressure of gas in each part is P°. Calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.
i wan hlp 2 undrstnd calorimetry
please tell us about the specific problem or topic that you are not able to understand. We will you detail explanation about that.
how did the specific heat capacity of water come in q-2???
You need to learn the specific heat capacity of water. There will be many questions where you will not be given Specific heat of water but still they are using it.
actually, if u refer example given on page 43 u would see that they added the water equivalent and the mass.if we do the same way we and take specific heat as pre described 4180,we get the answer as 449.3 which is the answer at the back of the book..........just wanted to know why have we added it...........................
Here do not get confused with the word water equivalent.
You can understand in this way that the calorimeter is also made up of some material. The heat which is given to the water, some heat will also be taken up by the calorimeter or we can say that some heat will be lost to the calorimeter material. so, we see that how much heat is taken by calorimeter mterial and calculate the equivalent mass of water that would take the same heat to change its temperature as that of the calorimeter.
That's why we are adding it with the given mass of water.
but water equivalent is explained on pg.40 of the book........which i have not understood
Here i am going to explain you about the water equivalent.
First of all the question arises from where this water equivalent is coming.
Now, see that there is calorimeter which is vessel made up of copper. Inside the calorimeter we have kept some cold water.
Now when this calorimeter comes in contact with some hot body then there will be transferring of heat energy, the whole system will try to attain equilibrium temperature in which the temperature of hot body will decrease and the temperature of cold body or the calorimeter containing water will increase. so here the heat from the hot body is entering the calorimeter. Let us assume that the temperature of colt water is increasing from T1 to T2. This means that the temperature of the container or the calorimeter in which the water is kept will also increase from T1 to T2. so, here we need to realize that some heat is used is increasing the temperature of the container.
Now I am going to explain that from where water equivalent is coming.
For this you need to understand the heat capacity.
Heat capacity of the material is the amount of heat required or released to change the temperature of the material of mass m by 1˚. so, the calorimeter is made of the copper whose temperature is increased from T1 to T2.
so, Total heat use by the container = ms(T2 - T1)
so, water equivalent is the mass of the water having same heat capacity as that of the container.
mˊc(T2 -T1) = ms(T2 - T1)
c = specific heat capacity of water
mˊ = ms/c
where mˊ = water equivalent
Now the question arises that why we are adding it with the given mass of water. This is so because we know that
Heat lost by hot body = Heat gained by calorimeter
= Heat gained by water filled in container + heat gained by container material
heat lost = mc(T2-T1) + heat gained by container ................eq(1)
where m = mass of water
c = specific heat
In above discussion i have shown you that Heat gained by container = Heat gained by water equivalent
SO, eq(1) changes to
Heat lost = mc(T2-T1) + mˊc(T2-T1) = (m+mˊ)c(T2-T1)
where m = mass of water
mˊ = water equivalent
wow man, you explain the whole stuff as well??
PLESE HELP IN SOLVING THIS PROBLEM?????
A Cooper Cube Of mass 200g slides down a rough incline plane of inclination 37° at a constant speed.Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in temp. of the block as it slides down 60cm. Specific heat capacity of cooper = 420J/kg-K
Here a cube is moving on the rough incline plane. so, we will get two components of weight. These are -
1. mg cos Ɵ perpendicular to the inclined plane.
2. mg sin Ɵ parallel to the inclined plane.
Here in this question it is given that the surface is rough, so friction force is acting on the body.
Also it is given that the body is moving with constant velocity. This means that acceleration is zero or the force acting along the inclined plane is zero. This means that friction force is balancing mg sin Ɵ.
friction force = mg sin Ɵ
As cube slides down 60cm. Then work done by the friction force or we can say loss of mechanical energy = friction force x distance
work done = mg sin Ɵ x 0.60 J
where m = 0.2 kg
Ɵ = 37°
In the question it is given that loss of mechanical energy goes into the copper as thermal energy.
so, Thermal energy = Loss of mechanical Energy
Thermal Energy = mcT
where c = heat capacity of cooper
T = increase in temperature
mcT = mg sin Ɵ x 0.60
T = (g sin Ɵ x 0.60)/c
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how to do q-4 (b)??
In the first part we see that the water and ice attains equilibruim.
The heat released by the water will be used by the ice to change into water but the heat released by water is not adequate to change the whole ice into water. Let us assume mass m of the ice changes into water.
m x latent heat of fusion = Heat released by water
Now calculate m
answer of ques 15 in h.c. verma is 3000 and by ur method it is coming 333.33 J.....is this the correct solution?
It may be possible that answer given in the book.
Still i am telling you how we solved this question. Please go to page no 148 in HC verma part i.
There you will see the formula for loss of kinetic energy in in elastic collision.
And we know that loss kinetic energy = thermal energy developed.
You may check yourself.
please calorimetry ka pura solution de
pura solution to diya hua hai. ab kya chahiye?
Can you explains me questions for short answer, question no. 2 , 4, 5, 6 and 10 .
Can you please explain me questions for short answer questions no. 2 , 4, 10.
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it is very helpful site
please explain me question no.4
please explain me question no. 5
as we know the water is cooling down so it is transferring its heat to atmosphere hence we need to find total heat tranfer and then rate of flow of heat secondly.
and as rate of flow of heat is = Q/T
thus T=total heat /rate of flow of heat
please explain 4Q PART A why the tempertature chages to 0 degrees of 200 ml
as the minimum temperature attained can be freezing point that is 0 when energy absorbed by ice to melt is lesser than the energy of ice thus whole cube will not melt.
i m new to this site. i am unable to get full answers of calorimetry. please help. plus i dont know if the comments i submit are able to reach to u
we try to answer al questions asked here.
but in case we miss your question you are free to ask any time again.
finally got it. thanks a lot
please help me sove the first problem
as heat lost by one object = heat gain by other object.
now heat gain by water is
heat of iron+heat of aluminum
mass of iron* change in temp * specific heat of iron + mass of aluminum * change in temp * specific heat of aluminum
sir, Is heat a conserved quantity?
no heat is not conserved its transient energy.
sir, after solving the que.6,i got the temperature 85.5*C but its different from given ans...
Plz give solution
calculations in given solution is right plz check if you have done any mistake.
sir, plz give the solution of Q.16 of exercise.
the question is that we calculate mechanical energy that later changes into 40% of heat
so first we get mechanical energy then we gets its 40% and then we have change in temperature
from change in heat=specific heat *mass*change in temperature
so we can find change in temperature
plz, show how i get mechanical energy
here we only have P.E.
so we can find P.E. in each case
thus we will have mechanical loss by subtracting it.
plz give the solution of Q.18
loss in energy=potential energy
now heat gained by both matters
mass of block*specific heat of block* change in temperature +
mass of water* specific heat of water* change in temperature= 4.707
thus change in temperature *( Mb * Cb + Mw *Cw)=4.707
I get formulas of physics...
there are handbooks for formulas of physics in market you can purchase them
or you better study them in books and make notes.
in ch-25 objective-I Q-1,
specific heat (s)=q/m(dt)
where q=heat ; m=mass ; dt=change in temp
hence the answer should be (a),(b),(c)
this is change in heat. . .
specific heat means specifically for a particular material hence it does depends upon material.
Thanks for these precious solutions and better too.
i want to know the true meaning of water equivalent
please share the whole line or paragraph you have read water equivalent only then i can explain true meaning
where i Get the solution of exersise of chapter 25
here you have solutions of exercise
In q7 what is the temperature at which thermal equillobrium is reached
dear it is not required for this instance.
as well we dont have all quantities to find equilibrium temp
So sir why it has taken that 1 kg water has taken from 0° to 100° . So this means that ice instantly melted at 0° and then melted water is taken from 0° to 100°. How can we say that water is taken to 100° without knowing equilibrium temp°
let us say the equilibrium temperature is 40' now what is use of it?
where do you think you would use it when we have calculated it.
Ok sir so if equillobrium temp is 40° then we must take heat gained by water (melted from ice) to ms(T2-T1) and change in temp must be 40° so I m having doubt how it can be 100° without water reaching till that temp??
Sir I got it!! The thermal equillobrium have be at 100°c as we know that all of the vapours is not going to be condensed and we know during condensation temp. Is const. So until whole of the vapours get condensed temp. Can't decrease!!!
in Q16 where did mass m go
Dear mass get cancelled all over . Don't see solution before solving it on your own dear!!!!
dear you can ask again if you want to..
sir please tell me ques 1of chapter 25 i used final temp as Q but i got wrong
dear where did you use temp as Q
Q is sign of heat you cant use it in place of temp
A calorimeter contains some ice 2.1kg of heat is required to increase the temp of the calorimeter and its contents from 270K and 69.72kg of heat is required from 272K to 274K.Find the mass
Inside an oven,there is high temp.we can put hand in it unless we touch anything.but since air inside is also of high temp, why is not hand burnt,plz provide me answer
heat is transfered through medium if medium is densed it will transfer more heat,
and if its gas it will transfer lesser heat.
hence when you touch hot solid at same temp you get more heat on your hand
but at same temp of air you get lesser heat.
If its like that then why steam at 100° is more dangerous than water at same temp. Probably the explanation is gas molecules have greater internal energy due to there random motion . So why is not like in the case of oven as temp inside it is also high.
its that steam are water vapors and it has latent heat in it.
but dry air is bad conductor of heat so it does not burn your hand at same instance but if you keep it in for more time you ll get burnt.
but if you touch a metal it will transfer heat at same instance and you ll get burnt.
Sir , i have a question . Force between two identical charges placed at a distance R in vacuum is F now a slab of dielectric constant 4 is inserted between two charges if the thickness of slab is R/2 then the force will become between th two charge ==
Sorry for puting it here!!