32 + Interview Questions in CALORIMETRY-AND-THERMAL-EXPANSION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Calculate amount of heat required to convert `1kg` steam form `100^(@)C to 200^(@)C` steam
Answer» Heat required `= 1 xx (1)/(2) xx 100 = 50 kcal`

Discussion

2.	A bullte of mass `10 gm` in moving with speed `400 m//s`. Find its kinetic energy in calories?
Answer» `Deltak = (1)/(2) xx (10)/(1000) xx 400 xx 400 = 800 rArr (800)/(4.2) = 191.11 Cal`.

Discussion

3.	`1kg` ice at `-10^(@)C` is mixed with `1kg` water at `100^(@)C`. Then final equilirium temperature and mixture content.
Answer» Heat taken by `1kg` Ice = Heat given by `1kg` water `1 xx (1)/(2)xx 10 +1 xx 80 +1 xx T = 1 xx (100 -T)` `85 -100 - 2T rArr 2T = 15` `theta = (15)/(2) = 7.5^(@)C`, water

Discussion

4.	`1kg` ice at `-10^(@)` is mixed with `1kg` water at `50^(@)C`. The final equilibrium temperature and mixture content.
Answer» Heat taken by ice `5= Kcal + 80 Kcal = 85 Kcal` Heat given by water `= 1 xx 1 xx 50 = 50 Kcal` Heat `gt` Heat given so, ice will not complete melt let m g ice melt then `1 xx (1)/(2) xx 10 +80 m = 50` `80 m = 45 rArr m = (45)/(80)` Content of mixture `{:{("water"(1+(45)/(80))kg),("ice"(1-(45)/(80))kg):}}` and temperature is `0^(@)C`

Discussion

5.	Explain, what is meant by the coefficients of linear `(alpha)`, superficial `(beta)` and cubical expansion `(gamma)` of a solid. Gien their units. Find the relatisphip between them.
Answer» Correct Answer - `alpha : beta : gamma = 1 : 2: 3` `SI` unit for `alpha, beta, gamma` are `K^(-1)` or `.^(@)C^(-1)`

Discussion

6.	Write the `S.I`. unit of temperature.
Answer» Correct Answer - Kelvin (symbol `:K)`

Discussion

7.	Is is possible to liquefy a gas at any temperature?
Answer» Correct Answer - No. Beyond critical temperature, it is not possible to liqueft the gas however large the pressure may be.

Discussion

8.	At what temperature do the Celsius and Fahrenheit readings have the same numerical value ?
Answer» Correct Answer - `-40^(@)C`.

Discussion

9.	When hot water is poured on a glass plate, it breaks because of
Answer» Correct Answer - When boiling water is poured into a thick tumbler, its inner surface expands. However, due to low thermal conductivity of glass, the expansion of outer surface of the tumbler is quite small. Due to uneven expansion of the outer and inner surfaces, the tumbler cracks.

Discussion

10.	A piece of ice falls from a height `h` so that it melts completely. Only one-quarter of the heat produced is absobed by the ice and all energy of ice gets converted into heat during its fall. The value of `h` is [Latent heat of ice is `3.4 xx 10^(5) J//kg` and `g = 10 N//kg`]
Answer» Correct Answer - `136 km`

Discussion

11.	A pendulum clock having copper rod keeos correct time at `20^0 C`. It gains 15 seconds per day if cooled to `0^@C`. Calculate the coefficient of linear expansion of copper.
Answer» `(15)/(24 xx 60 xx 60) = (1)/(2) alpha xx 20` `rArr alpha = (1)/(16 xx 3600) = 1.7 xx 10^(-5)//.^(@)C`

Discussion

12.	The gas thermometer are more sensitive than the liquid thermometers because gasesA. Statement-1 is True, Statement-2 is True, Statement-2 is correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is Flase, Statement-2 is True.
Answer» Correct Answer - A

Discussion

13.	A 0.60 kg sample of water and a sample of ice are placed in two compartmetnts A and B separated by a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice through the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and the ice are given in graph as functions of time t. Temperature of the compartments remain homogeneous during whole heat transfer process. Given specific heat of ice `=2100 J//kg-K`, specific heat of water `=4200 J//kg-K`, and latent heat of fusion of ice `=3.3xx10^5 J//kg`. The mass of the ice formed due to conversion from the water till thermal equilibrium is reached is equal toA. `0.12 kg`B. `0.15 kg`C. `0.25 kg`D. `0.40 kg`
Answer» Correct Answer - B

Discussion

14.	A simple seconds pendulum is constructed out of a very thin string of thermal coefficient of linear expansion `alpha = 20 xx 10^(-4)//^(@)C` and a heavy particle attached to one end. The free end of the string is suspended from the ceiling of an elevator at rest. the pendulum keeps correct time at `0^(@)C`. when the temperature rises to `50^(@)C`, the elevator operator of mass `60 kg` being a student of Physics accelerates the elevator vertically, to have the pendulum correct time. the apparent weight of the operator when the pendulum keeps correct time at `50^(@)C` is `("Take " g = 10m//s^(2)`)
Answer» Correct Answer - 66

Discussion

15.	A glass vessel measures exactly `10 cm xx 10 cm xx 10 cm` at `0^@C`. It is filled completely with mercury at this temperature. When the temperature is raised 10 `10^C, 1.6cm^3` of mercury overflows. Calculate of linear expansion of glass `= 6.5 xx 10^(-6 @)C^(-1)`.
Answer» `DeltaV = V_(Hg) - V_(v)` `1.6 = 10^(3) gamma_(l) xx 10 - 10^(3) xx3xx 6.5 xx 10^(-6) xx 10` `gamma_(L) = (1.6 + 0.195) xx 10^(-4) = 1.795 xx 10^(-4)//.^(@)C`

Discussion

16.	If percentage change in length is `1%` with change in temperature of a cuboid object `(l xx 2l xx3l)` then what is precentage change in its area and volume.
Answer» percentage change in length will change in temperature `= % l` `(Deltal)/(l) xx 100 = alpha Delta theta xx 100 = 1` change in area `rArr %A = (DeltaA)/(A) xx 100 = beta Delta theta rArr 2 (alpha Delta theta xx 100)` `% A = 2%` change in volume `% V = (DeltaV)/(V) xx 100 = V Delta theta xx 100 = 3 (alpha Delta theta xx 100)` `% V = 3%`.

Discussion

17.	The temperature of a metal ball is raised. Arrange the percentage change in volume, surface area and raidus in ascending order.
Answer» Correct Answer - `% R lt %A lt %V`

Discussion

18.	(a) The brass scale of a barometer gives correct reading at `0^(@)C` . Coefficient of linear expansion of brass is `2.0xx10^(-5)//^(@)C` . The barometer reads `75cm` at `27^(@)C` . What is the atmospheric pressure at `27^(@)C` ? (b) A barometer reads `75.0cm` on a steel scale. The room temperature is `30^(@)C` . The scale is correctly graduated for `0^(@)C ` . The coefficient of linear expansion of steel is `alpha=1.2xx10^(-5)//^(@)C` and the coefficient of volume expansion of mercury is `gamma=1.8xx10^(-4)//^(@)C` . Find the correct atmospheric pressure.
Answer» Correct Answer - `75.0405 cm` `h = h_(0) (1+alpha Delta theta)` `h = 75 {1+ 0.00002 xx 27}` `h = 75.0405 cm`

Discussion

19.	A uniform solid brass sphere is rotating with angular speed `omega_0` about a diameter. If its temperature is now increased by `100^@C`, what will be its new angular speed. (given`alpha_B=2.0xx10^-5per^@C`)A. `(omega_(0))/(1-0.002)`B. `(omega_(0))/(1+0.002)`C. `(omega_(0))/(1+0.004)`D. `(omega_(0))/(1-0.004)`
Answer» `I_(0) omega_(0) = I_(t) omegat` `Mr_(0)^(2) omega_(0) = Mr_(0)^(2) (1+2 alpha Delta T) omega_(t)` `omega_(t) = (omega_(0))/(1+0.004)`

Discussion

20.	A bar measured with a vernier caliper is found to be 180 mm long. The temperature during the measurement is `10^@C`. The measurement error will be if the scale of the vernier caliper has been graduated at a temeprature of `20^@C`. (`alpha=1.1xx10^(-5).^(@)C^(-1)`. Assume that the length of the bar does not change.)A. `1.98 xx 10^(-1) mm`B. `1.98 xx 10^(-2)mm`C. `1.98 xx 10^(-3)mm`D. `1.98 xx 10^(-4)mm`
Answer» True measurement =scale reading `[1 + alpha (theta - theta_(0))]` `= 180 [1- 10 xx 1.1 xx 10^(-5)]` error `= 180 - 180 [1-1.1 xx 10^(-4)] = 1.98 xx 10^(-2) mm`

Discussion

21.	Time taken by a 836 W heater to heat one litre of water from `10^@C to 40^@C` isA. `50 s`B. `100s`C. `150s`D. `200s`
Answer» Correct Answer - C Let time taken in boiling the water by the heater is t sec. Then `Q = ms Delta T` `(836)/(4.2) t = 1 xx 1000 (40^(@) - 10^(@))` `(836) /(4.2) t = 1000 xx 30 rArr t = (100 xx 30 xx 4.2)/(836) = 150 sec`

Discussion

22.	What is the principle of calorimetry.
Answer» Correct Answer - Heat lost by a substance is equal to heat gained by the other substance when they are brough in thermal contact.

Discussion

23.	`200 gm` water is filled in a calorimetry of negligible heat capacity. It is heated till its temperature is increase by `20^(@)C`. Find the heat supplied to the water.
Answer» `H = 200 xx 10^(-3) xx1 xx 20 = 4 Kcal` Heat supplied ` =4000 cal`.

Discussion

24.	In a container of negligible heat capacity, `200 gm` ice at `0^(@)C` and `100gm` steam at `100^(@)C` are added to `200 gm` of water that has temperature `55^(@)C`. Assume no heat is lost to the surrpundings and the pressure in the container is constant `1.0 atm`. (Latent heat of fusion of ice `=80 cal//gm`, Latent heat of vaporization of water `= 540 cal//gm`, Specific heat capacity of ice `= 0.5 cal//gm-K`, Specific heat capacity of water `=1 cal//gm-K)` What is the final temperature of the system ?A. `48^(@)C`B. `72^(@)C`C. `94^(@)C`D. `100^(@)C`
Answer» Correct Answer - D As steam has comparatively large amount of heat to provide in the form at latent heat we check what amount of heat is required by the water and ice to go up to `100^(@)C` that is `(m_(i)L +m_(i)S_(w)DeltaT)+ (m_(w).S_(w).DeltaT)` `= [(200 xx 80) +(200 xx1xx 100)] +(200 xx1xx 45)` `= 45,000cal` That is given by m mass of steam, then `m_(s)L = 45,000` `m_(s) = (45,000)/(540) = (500)/(6) = 83.3 gm` therefore `83.3 gm` steam converts into water of `100^(@)C`. Total water `= 200 +200 +83.3 = 483.3 gm` steam left `- 16.7 gm`.

Discussion

25.	In a container of negligible heat capacity `200` gm ice at `0^(@)`C and 100 gm steam at `100^(@)`C are added to 200 gm of water that has temperature `55^(@)`C. Assume no heat is lost to the surroundings and the pressure in the container is constant of `1.0atm (L_(f)=80 cal//gm,L_(v)=540cal//gm, s_(w)=1 cal//gm^(@)C)` At the final temperature, mass of the total water present in the system, isA. `472.6 gm`B. `483.3 gm`C. `483.6 gm`D. `500 gm`
Answer» Correct Answer - B As steam has comparatively large amount of heat to provide in the form at latent heat we check what amount of heat is required by the water and ice to go up to `100^(@)C` that is `(m_(i)L +m_(i)S_(w)DeltaT)+(m_(w).S_(w).DeltaT)` `= [(200 xx 80) +(200 xx1xx 100)] +(200 xx1xx 45)` `= 45,000cal` That is given by m mass of steam, then `m_(s)L = 45,000` `m_(s) = (45,000)/(540) = (500)/(6) = 83.3 gm` therefore `83.3 gm` steam converts into water of `100^(@)C`. Total water `= 200 +200 +83.3 = 483.3 gm` steam left `- 16.7 gm`.

Discussion

26.	`20 gm` ice at `-10^(@)C` is mixed with `m gm` steam at `100^(@)C`. The minimum value of `m` so that finally all ice and steam converts into water is: `("Use " s_("ice") = 0.5 "cal gm"^(@)C, S_("water") = 1 cal//gm^(@)C, L`) (melting) `= 80 cal// gm` and `L ("vaporization") = 540 cal//gm`)A. `(85)/(32)gm`B. `(85)/(64)gm`C. `(32)/(85)gm`D. `(64)/(85)gm`
Answer» Correct Answer - A For minimum value of `m`, the final temperature of the mixture must be `0^(@)C`. `:. 20 xx (1)/(2) xx 10 +20 xx 80 = m 540 +m.1. 100` `:. M = (1700)/(640) = (85)/(32) gm`.

Discussion

27.	A thermally insulated, closed copper vessel contains water at `15^@C`. When the vessel is shaken vigorously for 15 minuts, the temperature rises to `17^@C`. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are `420(J)/(kg-K)` and `4200(J)/(kg-K)` resprectively. Neglect any thermal expansion. (a) How much heat is transferred to the liquid vessel system? (b) How much work has been doen on this system? (c ) How much is the increase in internal energy of the system?
Answer» Correct Answer - (a) zero (b) `1764 J` (c ) `1764 J` (a), Zero because there is no heat transfter form outside. (b) `W = 100 xx 10^(-3) xx 420 xx2xx 200 xx10^(-3) xx4200 xx2 = 1764 J`. (c ) `DeltaU = 1764 J`

Discussion

28.	`420 J` of energy supplied to `10g` of water will rises its temperature by
Answer» `(420 xx 10^(-3))/(4.20) = 10 xx 10^(-3) xx1xx Deltat = 10^(@)C`

Discussion

29.	We have a hollow sphere and a solid sphere equal radii and of the same material. They are heated to raise their temperature be equal amounts. How will the cange in their volumes, due to volume expansions, be related? Consider two cases (i) hollow sphere is filled with air, (ii) there is vaccum inside the hollow sphere.
Answer» Correct Answer - (i) hollow sphere `gt` solid sphere (ii) hollow sphere = solid sphere When hollow sphere is filled air, expansion will be due to pressure of the gas also.

Discussion

30.	Heat required to raise the temperature of one gram of water through `1^@C` is
Answer» heat required `= 1 xx 10^(-3) xx1 xx1 = 1 xx 10^(-3) kcal = 1 cal`

Discussion

31.	If rod is initially compressed by `Deltal` length then what is the strain on the rod when the temperature (a) is increased by `Delta theta` (b) is decreased by `Delta theta`
Answer» (a) Strain `=(Deltal)/(l)+ alpha Delta theta` (b) Strain `=\|(Deltal)/(l)-alpha Delta theta\|`

Discussion

32.	The figure shows three temperature scales with the freezing and boiling points of water indicated. (a) Rank the size of a degree on these scales, greatest first. (b) Rank the following temperatures, highest first `:50^(@)X,50^(@)W` and `50^(@)Y`.
Answer» Correct Answer - (a) All tie (b) `50^(@)X, 50^(@)Y, 50^(@)W`.

Discussion

Explore topic-wise InterviewSolutions in .

Calculate amount of heat required to convert `1kg` steam form `100^(@)C to 200^(@)C` steam

A bullte of mass `10 gm` in moving with speed `400 m//s`. Find its kinetic energy in calories?

`1kg` ice at `-10^(@)C` is mixed with `1kg` water at `100^(@)C`. Then final equilirium temperature and mixture content.

`1kg` ice at `-10^(@)` is mixed with `1kg` water at `50^(@)C`. The final equilibrium temperature and mixture content.

Explain, what is meant by the coefficients of linear `(alpha)`, superficial `(beta)` and cubical expansion `(gamma)` of a solid. Gien their units. Find the relatisphip between them.

Write the `S.I`. unit of temperature.

Is is possible to liquefy a gas at any temperature?

At what temperature do the Celsius and Fahrenheit readings have the same numerical value ?

When hot water is poured on a glass plate, it breaks because of

A piece of ice falls from a height `h` so that it melts completely. Only one-quarter of the heat produced is absobed by the ice and all energy of ice gets converted into heat during its fall. The value of `h` is [Latent heat of ice is `3.4 xx 10^(5) J//kg` and `g = 10 N//kg`]

A pendulum clock having copper rod keeos correct time at `20^0 C`. It gains 15 seconds per day if cooled to `0^@C`. Calculate the coefficient of linear expansion of copper.

A glass vessel measures exactly `10 cm xx 10 cm xx 10 cm` at `0^@C`. It is filled completely with mercury at this temperature. When the temperature is raised 10 `10^C, 1.6cm^3` of mercury overflows. Calculate of linear expansion of glass `= 6.5 xx 10^(-6 @)C^(-1)`.

If percentage change in length is `1%` with change in temperature of a cuboid object `(l xx 2l xx3l)` then what is precentage change in its area and volume.

The temperature of a metal ball is raised. Arrange the percentage change in volume, surface area and raidus in ascending order.

Time taken by a 836 W heater to heat one litre of water from `10^@C to 40^@C` isA. `50 s`B. `100s`C. `150s`D. `200s`

What is the principle of calorimetry.

`200 gm` water is filled in a calorimetry of negligible heat capacity. It is heated till its temperature is increase by `20^(@)C`. Find the heat supplied to the water.

`420 J` of energy supplied to `10g` of water will rises its temperature by

Heat required to raise the temperature of one gram of water through `1^@C` is

If rod is initially compressed by `Deltal` length then what is the strain on the rod when the temperature (a) is increased by `Delta theta` (b) is decreased by `Delta theta`

The figure shows three temperature scales with the freezing and boiling points of water indicated. (a) Rank the size of a degree on these scales, greatest first. (b) Rank the following temperatures, highest first `:50^(@)X,50^(@)W` and `50^(@)Y`.