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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
The apparent corfficient of expansion of a liquid when heated, filled in vessel A and B of identical volumes, is found to be `gamma_1` and `gamma_2`, respectively. If `alpha_1` be the linear expansivity of A then that of B will beA. `((gamma_1-gamma_2))/(3)-alpha_1`B. `((gamma_2-gamma_1))/(3)+alpha_1`C. `((gamma_2-gamma_1))/(3)-alpha_1`D. `((gamma_1-gamma_2))/(3)+alpha_1` |
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Answer» Correct Answer - B Let V be the volume of the liquid and `DeltaT` the rise in temperature Since apparent expansion`=`true expansion`-`expansion of vessel `Vgamma_1DeltaT=VgammaDeltaT+V(3alpha_1)DeltaT` or `gamma_1=gamma+3alpha_1` .(i) and `gamma_2=gamma+3alpha_2` .(ii) Where `gamma` is the coefficient of real expansion of the liquid. Substracting Eq. (ii) from Eq. (i) `gamma_1-gamma_2=3(alpha_1-alpha_2)` or `alpha_1-alpha_2=((gamma_1-gamma_2))/(3)` or, `alpha_2=[(gamma_1-gamma_2)//3]+alpha_1` Hence, correct option is (b). |
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| 302. |
Statement-1: Gas thermometers are more sensitive than liquid thermometers. Statement-2: Coefficient of thermal expansion of gases is more than liquid.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A Gas thermometers are more sensitive then liquid thermometers because coefficient of thermal expansion of gases are more than liquids |
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| 303. |
The absolute coefficient of expansion of a liquid is 7 times that the volume coefficient of expansion of the vessel. Then the ratio of absolute and apparent expansion of the liquid isA. `(1)/(7)`B. `(7)/(6)`C. `(6)/(7)`D. none of these |
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Answer» Correct Answer - B Apparent coefficient of volume expansion `gamma_("app")=gamma_L-gamma_S=7gamma_S-gamma_S=6gamma_S` Ratio of absolute and apparent expansion of liquid `(gamma_L)/(gamma_("app"))=(7gamma_S)/(6gamma_S)=(7)/(6)` |
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| 304. |
A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from `100^@C` to `99^@C` in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from `100^@C` to `99^@C` will be approximatelyA. 25 sB. 50 sC. 200 sD. 400 s |
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Answer» Correct Answer - C `(dT)/(dt)=(eAsigma)/(mc)(T^4-T_0^4)=(e(6a^2)sigma)/((a^3xxrho)c)(T^4-t_0^4)` `implies` for the same fall in temperature time `dtpropa` `(dt_2)/(dt_1)=(a_2)/(a_1)=(2cm)/(1cm)impliesdt_2=2xxdt_1=2xx100s=200s` (As `A=6a^2` and `m=Vxxrho=a_3xxrho`) |
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| 305. |
A cube of mass 1 kg and volume `125cm^2` is placed in an evacuated chamber at `27^@C`. Initially temperature of block is `227^@C`. Assume block behaves like a block body, find the rate of cooling of block if specific heat of the material of block is `400 J//kg-K`. |
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Answer» In this case ther rate of loss of heat by the block is given as `(dQ)/(dt)=sigmaA(T^$-T_0^4)` `=5.67xx10^-8xx150xx10^-4xx[(500)^4-(300)^4]` (surface area of cube is `6a^2=150cm^2`) If `(dT)/(dt)` is rate of cooling of block then we have `(dQ)/(dt)=ms(dT)/(dt)` or `(dT)/(dt)=(1)/(ms)(dQ)/(dt)=(1)/(1xx400)xx38.556=0.115^@C//s` |
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| 306. |
Four cylindrical rods of same material with length and radius (l,r),(2l,r),(2l,r) and (l,2r) are connected between two reservoirs at `0^@C` and `100^@C`. Find the ratio of the maximum to minimum rate of cunduction in them. |
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Answer» Rate of conduction `Rprop(r^2)/(l)` The ratio of conduction in them is `(r^2)/(l):(r^2)/(2l):(2r^2)/(l):(4r^2)/(l)`, i.e. `2:1:4:8` So, the ratio of maximum to minimum conduction rate is `8:1` |
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| 307. |
2 kg of ice at `-20^@C` is mixed with 5 kg of water at `20^@C` in an insulating vessel having a negligible heat capacity. Calculate the final mass of water (in kg) remaining in the container. |
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Answer» Energy with 5 kg of `H_2O` at `20^@C` to become ice at `0^@C` `E_1=5000xx1xx20=100000cal` Energy to raise the temperature of 2 kg ice from `-20^@C` to `0^@C` `E_1=5000xx0.5xx20=20000cal` `(E_1-E_2)=80000cal` is available to melt ice at `0^@C`. So only 1000 g or 1 kg of ice would have melt. So, the amount of water available`1+5=6kg` |
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| 308. |
`50 g` of steam at `100^@ C` is passed into `250 g` of at `0^@ C`. Find the resultant temperature (if latent heat of steam is `540 cal//g`, latent heat of ice is `80 cal//g` and specific heat of water is `1 cal//g -.^@ C`).A. `40^@ C`B. `30^@ C`C. `20^@ C`D. `10^@ C` |
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Answer» Correct Answer - A Heat lost by steam = Heat gained by ice. `m_(steam) L_(v)+m_(s) xxS_(w)(100^@-theta^@)` `=m_(w)L_(f) +m_(w)S_(w) (theta^@ - 0)`. |
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| 309. |
2 kg of ice at `-15^@C` is mixed with 2.5 kg of water at `26^@C` in an insulating container. If the specific heat capacities of ice and water are `0.5 cal//g^@C` and `1 cal//g^@C`, find the amount of water present in the container? (in kg nearest integer) |
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Answer» Energy relesased by water from `25^@C` to `0^@C` `=2500xx1xx25=62500kcal` Energy to bring ice to `0^@C` `=2000xx(1)/(2)xx15=15000cal` Energy used to melt ice of m gram `=m80cal` Ice melt `m=((62500-15000)/(80))=593.75g` So, mass of water `=(2500+593.75)g` `=3093.75gcong3kg` |
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| 310. |
`1 g` steam at `100^@C` is passed in an insulating vessel having `1 g` ice at `0^@C`. Find the equilibrium composition of the mixture. (Neglecting heat capacity of the vessel). |
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Answer» Available heat from steam `mL = 1 xx 540 = 540 cal` Heat required for melting of ice and to rise its temperature to `100^@ C = m_(ice) L_(ice) + m_(water) S_(water) Delta theta` =`(1 xx 80) +[1 xx 1 xx(100 - 0)] = 180 cal` Let `m` be the mass of steam condensed, then `m xx 540 = 180 rArr m = (180)/(540) = (1)/(2) g` Final contents : Water = `1 + (1)/(3) = (4)/(3) g`, steam `= 1 - (1)/(3) = (2)/(3) g`. |
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| 311. |
The following graph represents change of state of `1` gram of ice at `-20^@C`. Find the net heat required to convert ice into steam at `100^@ C`. `S_(ice) = 0.53 cal//g -^@ C` . |
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Answer» In the figure : `a to b` : temperature of ice increases until its reaches its melting point `0^@ C` `Q_(1)=mS_(ice)[0-(-20)] =(1)(0.53)(20)=10.6 cal` `b to c` : Temperature remains constabt until all the ice has melted. `Q_(2) = mL_(f) = (1)(80) = 80 cal` `c to d` : Temperature of water againm rises until it reaches its boiling point `100^@C` `Q_(3) = mS_(water) [100-0] = (1)(1.0)(100) = 100 cal` `d to e` : Temperature is again constant until all the water is transformed into the vapour phase `Q_(4) = mL_(v) = (1)(530) = 539 cal` Thus, the net heat required to convert `1 g` of ice at `-20^@ C` into steam at `100^@ C` is `Q = Q_(1) + Q_(2) + Q_(3) + Q_(4) = 729.6 cal`. |
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| 312. |
`30 kg` ice at `0^@C` and `20 g` of steam at `100^@C` are mixed. The composition of the resultant mixture isA. `40 g` of water and `10 g` steam at `100^@C`B. `10g` of ice and `40 g` of water at `0^@C`C. `50g` of water at `100^@C`D. `35 g` of water and `15 g` of steam at `100^@ C` |
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Answer» Correct Answer - A `m_(s)L_(s) =(m_(ice)L_(ice) + m_(ice) s Delta theta)` Where `m_(s)` = mass f steam condensed to rise temperature of ice to `100^@C` water. |
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| 313. |
Heat is required to change 1 kg of ice at `-20^@C` into steam. `Q_1` is the heat needed to warm the ice from `-20^@C` to `0^@C`, `Q_2` is the heat needed to melt the ice, `Q_3` is the heat needed to warm the water from `0^@C` to `100^@C` and `Q_4` is the heat needed to vapourize the water. ThenA. `Q_4gtQ_3gtQ_2gtQ_1`B. `Q_4gtQ_3gtQ_1gtQ_2`C. `Q_4gtQ_2gtQ_3gtQ_1`D. `Q_4gtQ_2gtQ_1gtQ_3` |
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Answer» Correct Answer - A `Q_4=mL_W=540cal` `Q_3=ms_W(100-0)=100cal` `Q_2=mL_("ice")=80cal` `Q_1=ms_("ice")(20-0)=20cal` `Q_(4) lt Q_(3) lt Q_(2) lt Q_(1)` |
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| 314. |
An electrically heated coil is immersed in a calorimeter containing 360 g of water at `10^@C`. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and coil is 40 g. The temperature of water after 10 min isA. `4.214^@C`B. `42.14^@C`C. `30^@C`D. none of these |
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Answer» Correct Answer - B Energy supplied by the heater to the system in 10 min `Q_1=Pxxt90 J//s xx10xx60s` `=54000J=(54000)/(4.2)cal=12857cal` Now if `theta` is the final temeprature of the system, energy absorbed by it to change its temperature from `10^@C` to `theta^@C` is `Q_2=(msDeltaT)_("water")+(msDeltaT)_("coil + calorimeter")` `=360xx1xx(theta-10)+40(theta-10)` `=400(theta-10)` According to problem, `Q_1=Q_2` So `12857=400(theta-10)` or `theta=42.14^@C` |
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| 315. |
An immersion heater, in an insulated vessel of negligible heat capacity brings 10 g of water to the boiling point from `16^@C` in 7 min. Then Q. The water is replaced by 200 g of alcohol, which is heated from `16^@C` to the boiling point of `78^@C` in 6 min 12 s whereas 30 g are vapourized in 5 min 6 s. The specific heat of alcohol isA. `0.6 J//kg^@C`B. `0.6 cal//g^@C`C. `0.6 cal//kg^@C`D. `6 J//kg^@C` |
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Answer» Correct Answer - B With 200 g of alcohol in the vessel, the temperature Quantity of heat absorbed by alcohol `Q_a=C_am_a(theta_f-theta_i)` `=C_axx(200g)(78-16)^@C` Heat given by heater `Q_a=Pt_2=(20(cal)/(s))xx372s` Specific heat of alcohol `C_a=(20xx372)/(200xx62)=0.6(cal)/(g^@C` `=(0.6xx4.186J)/(10^-3kg^@C)=2.5xx10^3(J)/(kg^@C)` `=2.5(K)/(g^@C)` |
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| 316. |
An immersion heater, in an insulated vessel of negligible heat capacity brings 10 g of water to the boiling point from `16^@C` in 7 min. Then Q. Power of heat is nearlyA. `8.4xx10^3`B. 84 WC. `8.4xx10^3(cal)/(s)`D. 20 W |
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Answer» Correct Answer - B In 7 min, temperature of 100 g of water is raised by `(1000-16^@C)=84^@C`. The amount of the heat provided by heater `Q_W=C_Wm_WDeltaT=((1cal)/(g^@C))(100g)(84^@C)` `=8.4xx10^3cal=(8.4xx10^3xx4.186)J` `=3.5xx10^4J` Power of heater `=(Q_W)/(t_1)` `=(8.4xx10^3cal)/((7xx60)s)=20(cal)/(s)` `=(20xx4.18)(J)/(s)=83.6Wapprox84W` |
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| 317. |
An electric heater whose power is 54 W is immersed in `650cm^3` water in a calorimeter. In 3 min the water is heated by `3.4^@C`. What part of the energy of the heater passes out of the calorimeter in the form of radiant energy? |
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Answer» Energy suupplied by heater`=Pt`. `=54xx3xx60=9720J` Energy absorbed by water `=(650xx10^-3)xx4200xx3.4=9282J` Therefore, energy that passes out in the form of radiant heat `=9720-9282=438J` Percentage loss`=(438)/(9720)xx100=4.5%` |
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| 318. |
Suppose an astronaut on the surface of the moon took some water at about `20^@C` out of his thermos and poured it into a glass bbeaker. What would happen to the water? |
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Answer» The external pressure being almost zero, some of the water would evaporate quickly. The evaporating water would take its latent heat of vapourization from the remaining water. Hence, the remaining water would first cool to `0^@C` and then freeze. Thus, ultimately, a part of the water would evaporate and the rest would freeze. This establishes the condition that a liquid cannot exist in an extended vacuum. |
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| 319. |
It takes 10 in for an electric kettle to theat a certain quantity of water from `0^@C` to `100^@C`. It takes 54 min to convert this water `100^@C` into steam Then latent heat of steam isA. 80 cal/gB. 540 cal/kgC. 540 cal/gD. 80 cal/kg |
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Answer» Correct Answer - C Let m be the mass of water. Quantity of heat absorbed by water in 10 min. `=msDeltaT=mxx1xx100=100m` (in calories) Quantity of heat absorbed by water in 54 min. `=(100mxx54)/(10)` Quantity of heat required to convert water into steam`=mL` Hence, `=(100mxx54)/(10)=mL` or `L=540 cal//g`. |
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| 320. |
Explain the following :Cold water is poured on the burns caused on the skin by some hot object. |
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Answer» Water has high sp. heat capacity and can remove more heat from the bums caused on the skin by hot object and releives of the pain. |
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| 321. |
By imparting heat to a body its temperature rises by 15° C. what is the corresponding rise in temperature on kelvin scale? |
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Answer» The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales. Therefore, the corresponding rise in temperature on the Kelvin scale will be 15K. |
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| 322. |
1.0 kg of water is container in a 1.25 kW kettle Calculate the time taken for the temperature of water to rise from 25° C to its boiling point 100°C. Specific heat capacity of water = 4.2 J g-1K-1 |
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Answer» Mass of water = 1000 g. Change in temperature=100oC − 25 oC = 75 oC (or 75K) Amount of heat energy required to raise temperature = 1000 × 4.2 × 75 = 315000 J. Time taken to raise the temperature, T = 4 min 12 seconds. |
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| 323. |
`10` litres of hot water at `70^@C` is mixed with an equal volume of cold water at `20^@ C`. Find the resultant temperature of the water. (Specific heat of water `= 4200 J//kg -K`). |
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Answer» Resultant temperature, `theta = (m_(1)S_(1)theta_(1)+m_(2)S_(2)theta_(2))/(m_(1)S_(1)+m_(2)S_(2))` Here, `m_(1) = m_(2) = 10 kg` (since mass = `1` litre of water is `1 kg`) `theta_(1) = 70^@ C, theta_(2) = 20^@ C` and `S_(1) = S_(2) = 4200 J//kg -K` `theta = (10xx4200 xx 70 +10 xx 4200 xx 20)/(10 xx4200 +10xx 4200) = 45^@ C`. |
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| 324. |
What is meant by Specific heat capacity of a substance? |
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Answer» Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of substance through 1C. It is denoted by C. Q = mcθ c = Q/mθ |
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| 325. |
Explain why water is used in hot water bottles for fomentation and also as a universal coolant. |
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Answer» For specific heat capacity of water being very high i.e. 4200JKg-1k-1, water extracts more heat from hot surrounding and loses it very slowly and acts as effective coolant. |
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| 326. |
A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at `40 .^(@)C`. When m gram of ice at `-10^(@)C` is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be `20^(@)C`. It is known that specific heat capacity of the liquid changes with temperature as `S = (1+(theta)/(500)) cal g^(-1) .^(@)C^(-1)` where `theta` is temperature in `.^(@)C`. The specific heat capacity of ice, water and the calorimeter remains constant and values are `S_("ice") = 0.5 cal g^(-1) .^(@)C^(-1), S_("water") = 1.0 cal g^(-1) .^(@)C^(-1)` and latent heat of fusion of ice is `L_(f) = 80 cal g^(-1)`. Assume no heat loss to the surrounding and calculate the value of m. |
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Answer» Correct Answer - 12g |
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| 327. |
A calorimeter of water equivalent `10xx10^(-3)kg` is filled with `70xx10^(-3)kg` of a substance at its melting point. The substance is found to solidify completely in 21 minutes. A similar calorimeter containing `80xx10^(-3)kg` of water at the same temperature cools at the rate of `1.5^(@)C` per minute, the room temperature being the same in both cases. What is the latent heat of fusioin of the substance? (Specific heat capacity of water `=4200 Jkg^(-1) K^(-1)`) |
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Answer» Correct Answer - `170xx10^(3)Jkg^(-1)` |
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| 328. |
A calorimeter whose water equivalent is `5xx10^(-3) kg` is filled with `50xx10^(-3)` kg of water at `80^(@)C`. The temperature falls to `75^(@)C` in 4 minutes. When it is filled with `40xx10^(-3)kg` of another liquid, the time to fall through same range of temperature (from `80^(@)C` to `75^(@)C` ) is 130 seconds. Find the specific heat capacity of the liquid. |
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Answer» Correct Answer - `2603Jkg^(-1)K^(-1)` |
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| 329. |
A calorimeter of water equivalent `100` grams contains `200` grams of water at `10^@ C`. A solid of mass `500` grams at `45^@ C` is added to the calorimeter. If equilibrium temperature is `25^@ C` then, the specific heat of the solid is `("in cal"//g -.^@ C)`.A. `0.45`B. `0.1`C. `4.5`D. `0.01` |
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Answer» Correct Answer - A Heat lost by solid = Heat gained by calorimeter and water `m_(s)S_(s)(45^@-25^@)=(m_(c)S_(c)+m_(w)S_(w))(25^@ -10^@)`. |
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| 330. |
A calorimeter contains ice. Determine the heat capacitiy of the calorimeter if `2.1kJ` of heat is required to heat it togheter with its contents from `270^(@)K` to `270^(@)K`, and `69.72kJ` of heat of required to raise its temperature of `272K` to `274^(@)K`. (`L` of ice `=336xx10^(3)Jkg^(-1)`, specific heat capacity of ice `=2100kg^(-1)K^(-1)`) |
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Answer» Correct Answer - `630JK^(-1)` |
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| 331. |
A steam at `100^@C` is passed into `1 kg` of water contained in a calorimeter of water equivalent `0.2 kg` at `9^@C` till the temperature of the calorimeter and water in it is increased to `90^@C`. Find the mass of steam condensed in `kg(S_(w) = 1 cal//g^@C, & L_("steam") = 540 cal//g) (EAM = 14 E)`. |
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Answer» Let, `m` be the mass of the steam condensed. Mass of the steam passed into calorimeter, `m_(2) = 1 kg = 1000 g`. Water equivalent of calorimeter, `m_(1)S_(1) = 0.2 kg = 200 kg` `theta_(1)` = temperature of the steam `= 100^@ C` `theta_(2)` = temperature of the water `= 9^@C` `theta_(3)` = resultant temperature `= 90^@C` Heat lost = heat gained (calorimeter + water) `m[L_(steam) + S_(W) (theta_(1) -theta_(3))] = [m_(1)S_(1) + m_(2) S_(W)] (theta_(3) -theta_(2))` `m[540+1(100-90)] = [200 + 1000 xx 1](90-9)` `rArr m = 176 = 0.176 kg~~ 0.18 kg`. |
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| 332. |
The height of a mercury column measured with a brass scale, which is correct and equal to `H_0` at `0^@C`, is `H_1` at `t^@C`? The coefficient of linear expansion of brass is `alpha` and the coefficient of linear expansion of brass is `alpha` and the corfficient of volume expansion of mercury is `gamma`. Relate `H_0` and `H_1`. |
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Answer» Since the scale is correct at `0^@C` 1 cm at `0^@C` exactly 1 cm `:.` 1 cm at `t^@C` `=`exactly`(1+alphat)cm` `:. H_1cm` at `t^@C=H_1(1+alphat)cm` Since pressure is the same, namely, the atmosphere pressure. `rho_0gH_0=rho_tgH_1(1+alphat)` or `rho_1(1+gammat)H_0=H_1rho_1(1+alpha)` or `H_0 ~~ H_1(1+alpha)(1-gammat)` |
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| 333. |
At room temperature (`25^@C`) the length of a steel rod is measured using a brass centimetre scale. The measured length is 20 cm. If the scale is calibrated to read accurately at temperature `0^@C`, find the actual length of steel rod at room temperature. |
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Answer» The brass scale is calibrated to read accurately at `0^@C`. This means at `0^@C`, each division of scale has exact 1 cm length. Thus at thigher temperature the division length of scale will be more than 1 cm due to thermal expansion. Thus at higher temperature the scale reading for length measurement is not appropriate and as at higher temperature the division length is more, the length this scale reads will be lesser than the actual length to be measure. For illustration in this case the length of each division on brass scale at `25^@C` is `l_(1"div") =(1cm)[1+alpha_(br)(25-0)]` `=1+alpha_(br)(25)` It is given that at `25^@C` the length of steel rod measured is 20 cm, actually is in not 20 cm, it is 20 divisions on the brass scale. Now we can find the actual length of the steel rod at `25^@C` as `l_(25^@C)=(20cm)xxl_(1 "div")` `i_(act)=20[1+alpha_(br)(25)]` The above expression is a general relation using which you can find the actual lengths of the objects of which length are measured by a metallic scale at some temperature other than the graduation temperature of the scale. |
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| 334. |
A barometer with a brass scale reads 755 mm on a day when the temperatures is `25^@C`. If the scale is correctly graduated at `0^@C`, find the true pressure at `0^@C` (interms of height of Hg) given that the coefficient of linear expansion of brass is `18xx10^(-6)//K`. Coefficient of cubical expansion of mercury`=182xx10^(-6)//K`. |
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Answer» Given that 1 mm at `0^@C=1mm` 755 mm at `25^@C=755(1+18xx10^-6xx25)mm=755.34mm` Let P be the value of the atmospheric pressure. Then `P=755.34rho_(25)g=hrho_(0)g`, where `rho_(0)`,`rho_(25)` are densities of mercury at `0^@C` and `25^@C`, respectively. or `h=755.34xx(rho_(25))/(rho_(0))=755.34xx(rho_0)/(rho_0(1+182xx10^-6xx25))` or `h=751.19` mm |
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| 335. |
A steel rod is 3.000 cm at `25^@C`. A brass ring has an interior diameter of 2.992 cm at `25^@C`. At what common temperature will the ring just slide on to the rod? |
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Answer» `alpha_("steel")=12xx10^-6K^-1` and `alpha_("brass")=18xx10^-6K^-1` Let t be the required common temperature Then `:. DeltaT=t-25`. At the common temperature, both must have the same diameter. `D=3.000(1+12xx10^-6DeltaT)` `=2.992(1+18xx10^-6DeltaT)` `DeltaT=448^@Cimpliest=448+25` `=473^@C` |
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| 336. |
Steam at `100^@C` is passed into a calorimeter of water equivalent 10 g containing 74 cc of water and 10 g of ice at `0^@C`. If the temperature of the calorimeter and its contents rises to `5^@C`, calculate the amount of steam passed. Latent heat of steam`=540 kcal//kg`, latent heat of fusion`=80 kcal//kg`. |
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Answer» `(mxx540+mxx1xx95)`kcal `=(10xx10^-3)xx80+(94xx10^-3)xx1xx5` kcal Hence `m=2xx10^-3kg` or `2g`. |
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| 337. |
An iron rod and another of brass, both at `27^@C` differ in length by `10^-3`m. The coefficient of linear expansion for iron is `1.1xx10^(-5)//^(@)C` and for brass is `1.9xx10^(-5)//^(@)C`. The temperature at which both these rods will have the same length isA. `0^@C`B. `152^@C`C. `175^@C`D. Data is insufficient |
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Answer» Correct Answer - A The given data is normally isufficient to determine the result. Either the length of one of the rods at `0^@C` must be known or one has to assume that both the rods had the same length at `0^@C` which is not possible if they have the same length again at another temperature. Suppose the two rods had the same length L at `0^@C`. Then by the given problem `L[1+1.9xx10^-5xx27]-L(1+1.1xx10^-5xx27]=10^-3` `Lxx0.8xx27xx10^-5=10^-3` or `L=(10^-3)/(0.8xx27xx10^-5)m=4.63m` Hence `0^@C` is a possible choice and the rods had equal length of `4.63m` at `0^@C`. |
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| 338. |
A brass scale is graduated at `10^@C`. What is the true length of a zinc rod which measures 60.00 cm on this scale at `30^@C`? Coefficient of linear expansion of brass`=18xx10^-6K^-1`. |
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Answer» Since the scale is graduated at `10^@C`, 1 cm of the scale at `10^@C` = exactly 1 cm `:. 1` cm of the scale at `30^@C` = exactly `(1+18xx10^(-6) xx20)` cm `:.` 60 cm of the scale at `30^@C` = exactly `60.00(1+36xx10^-5)` `=60.02cm`. |
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| 339. |
A calorimeter takes `200 cal` of heat to rise its temperature through `10^@C`. Its water equivalent in `gm` isA. `2`B. `10`C. `20`D. `40` |
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Answer» Correct Answer - C `H = (Delta Q)/(Delta theta)`. |
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| 340. |
1 g ice at `0^@C` is placed in a calorimeter having 1 g water at `40^@C`. Find equilibrium temperature and final contents. Assuming heat capacity of calorimeter is negligible small. |
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Answer» The heat available on water `=mcDeltat=1xx80=80cal` The heat available on water `=mcDeltat=1=1xx1xx(40-0)` `=40cal` Threfore, entire heat of water is utilized to melt the ice and its temperature falls to `0^@C` ice is still at `0^@C`. So equilibrium temperature of contents remains `0^@C`. Let m is the amount of ice that melts by absorbing 40 cal heat. Then `mxx80=40` or `m=(1)/(2)g` Final contents: ice `=1-(1)/(2)=(1)/(2)g` Water`=1+(1)/(2)=(3)/(2)g` |
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| 341. |
A copper calorimeter has mass of 180 g and contains 450 g of water and 50 g of ice, all at `0^(@)C`. Dry steam is passed into the calorimeter until a certain temperature `(theta)` is reached. The mass of the calorimeter and its contents at the end of the experiment increased by 25 g. Assume no heat loss to the surrounding and take specific heat capacities of water and copper to be `4200 J kg^(-1) K^(-1)` and `390 J kg^(-1) K^(-1)`, respectively. Take specific latent heat of vaporization of water to be `3.36 xx 10^(5) J kg^(-1)` and `2.26 xx 10^(6) J kg^(-1)` respectively. (a) Find the final temperature `theta` (b) If steam enters into the system at a steady rate of `5 g min^(-1)`, plot the variation of temperature of the system till final temperature `theta` is attained. |
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Answer» Correct Answer - (a) `22^(@)C` |
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| 342. |
A brass rod and a lead rod each 80 cm long at `0^@C` are clamped together at one end with their free ends coinciding. The separatioin of free ends of the rods if the system is placed in a steam bath is (`alpha_("brass")=18xx10^(-6)//^(@)C` and `alpha_("lead")=28xx10^(-6)//^(@)C`)A. 0.2 mmB. 0.8 mmC. 1.4 mmD. 1.6 mm |
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Answer» Correct Answer - B The brass rod and the lead rod will suffer expansion when placed in steam bath. Length of brass rod at `100^@C` `L_("brass")=L_("brass")(1+alpha_("brass")DeltaT)=80[1+18xx10^-6xx100]` and the length of lead rod at `100^@C` `L_("lead")=L_("lead")(1+alpha_("lead")DeltaT)=80[1+28xx10^-6xx100]` Separation of free ends of the rods after heating `=L_("lead")-L_("brass")=80[28-18]xx10^-4=8xx10^-2cm=0.8` mm |
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| 343. |
In two experiments with a countinous flow calorimeter to determine the specific heat capacity of a liquid,an input power of 16 W produced a rise of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of liquid three times faster. Find the power lost (in W) to the surrounding in each case. |
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Answer» Let power lost to surrounding is `Q` `16-!=((dm)/(dt))S(10)` and `32-Q=3[((dm)/(dt))S(10)]` `implies(32-Q)/(16-Q)=3impliesQ=8W` |
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| 344. |
`4g` of steam is added to a mixture of `35g` of ice and `35g` of water in a copper calorimeter weighing `50g`. What is the final temperature? Specific heat capacity of copper `=420Jkg^(-1)K^(-1)L` (of ice) `=336xx10^(3)Jkg^(-1)` and `L` (of steam) `=2268xx10^(3)Jkg^(1)` ) |
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Answer» Correct Answer - `3xx10^(-3)` kg of ice and `67xx10^(-3)` kg of water at `0^(@)C` |
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| 345. |
A mass m of lead shot is placed at the bottom of a vertical cardboard cylinder that is 1.5 m long and closed at both ends. The cylinder is suddenly inverted so that the shot falls 1.5 m If this procces is repeated quickely 100 times, auuming no heat is dissipated or lost, the temperature of the shot will increase by (specific heat of lead`=0.03 cal//g^@C`) |
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Answer» Correct Answer - D Specific heat of lead`=0.03 kcal//kg^(@)C`. Gravitational potential energy is converted into thermal energy which is absorbed by the lead shot. If `T_0` is the rise in temperature of the shot, then `(100)[mgh]=(mxxsxxT)J` `T=(100xxgxxh)/(Js)=(100xx9.8xx1.5)/(0.03xx4.18xx10^3).^(@)C=11.3^@C` |
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| 346. |
A liquid of density `0.85 g//cm^(3)` flows through a calorimeter at the rate of `8.0 cm^(2)//s`. Heat is added by means of a 250 W electric heating coil and a temperature difference of `15^@C` is established in steady state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will beA. `0.6kcal//kgK`B. `0.3kcal//kgK`C. `0.5kcal//kgK`D. `0.4 kcal//kgK` |
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Answer» Correct Answer - A This is a problem on flow calorimeter used to measure specific heat of a liquid. Amount of heat supplied to the water per second by the heating coil`=Q_S=250J` `=(250)/(4186)kcal` The volume of liquid flowing out per second `=8.0cm^3=8xx10^6m^3` Mass of this liquid`=(0.85)xx1000xx8xx10^-6kg` Temperature rise of this mass of liquid`=15^@C` hence, `(250)/(4186)=mst=0.85xx8xx10^-3xxsxx15` hence, `s=(250xx10^3)/(4186xx0.85xx8xx15)=0.6 kcal//kgK` |
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| 347. |
The coefficient of thermal conductivity of copper, mercury and glass are respectively `K_(c), K_(m)` and `K_(g)` that `K_(c) gt K_(m) gt K_(g)`. If the same quantity of heat is to flow per second per unit of each and corresponding temperature gradients are `X_(c), X_(m)` and `X_(g)`, thenA. `X_x=X_m=X_g`B. `X_c gt X_m gt X_g`C. `X_c lt X_m lt X_g`D. `X_m lt X_c lt X_g` |
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Answer» Correct Answer - C `(dQ//dt)/(A)=K((Deltatheta)/(Deltax))implies` Rate of flow of heat per unit area = Thermal conductivity`xx`Temperature gradient Temperature gradient `(X)prop(1)/("Thermal conductivity(K)")` (As `(dQ//dt)/(A)=` constant) As `K_C gt K_m gt K_G`, therefore `X_(c) lt X_(m) lt X_(g)` |
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| 348. |
A refrigerator converts 1.3 kg of water at `20^(@)C` into ice at `-15^(@)C` in 1 hour. Calculate the effective power of the refrigerator. Specific latent heat of fusion of ice `= 3.4 xx 10^(5) J kg^(-1)` Specific heat capacity of water `= 4.2 xx 10^(3) J kg^(-1) K^(-1)` Specific heat capacity of ice `= 2.1 xx 10^(3) J kg^(-1) K^(-1)` |
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Answer» Correct Answer - `164.5 W` |
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| 349. |
A refrigerator converts 100g of water at 20℃ to ice at – 10℃ in 73.5 min. Calculate the average the rate of heat extraction in watt. The specific heat capacity of water is 4.2 J kg-1 K-1, Specific latent heat of ice is 336 J g-1 and the specific heat capacity of ice if 2.1 J kg-1 K-1. |
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Answer» Amount of heat released when 100g of water cools from 20o to 0oC = 100 × 20 × 4.2 = 8400J. Amount of heat released when 100g of water converts into ice at 0oC = 100 × 336 = 33600J. Amount of heat released when 100g of ice cools from 0oC to -10oC = 100 × 10 × 2.1 = 2100J. Total amount of heat = 8400 + 33600 + 2100 = 44100J. Time taken = 73.5min = 4410s. Average rate of heat extraction (power). P = E/t = 44100/4410 = 10W |
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| 350. |
The greater the mass of a body , the greater is its latent heat capacity. Is this true of false? |
| Answer» True. (Note that the specific latent heat capacity does not depend on mass.) | |