360 + Interview Questions in CALORIMETRY IN GENERAL AWARENESS Page 4 InterviewSolution

151.	Explain the formation of sea breeze.
Answer» Sea breeze : Blowing of cool air from sea towards land. During day time land gets heated up rapidly due to low sp. heat of land as compared to water. Pressure at land decreases. Hence air blows from sea (high pressure) towards land (low pressure).

Discussion

152.	Why is the weather in coastal regions moderate?
Answer» The climate near coastal regions moderate : The sp. heat capacity of water is very high or sp. heat capacity of land is much low as compared to water. As such land (or sand) gets cooled more rapidly as compared to water under similar conditions. Thus, a large difference in temperature is developed between the land and the sea, due to which cold air blows from land towards sea during night (i.e. land breeze) and during the day cold air blows from sea towards land (i.e. sand breeze). These make the climate near coastal region moderate.

Discussion

153.	The air temperature above coastal areas is profoundly infuenced by the large specific heat of water. One reason is that the energy released when `1m^3` of water cools by `1^@C` will raise the temperature of a much larger volume of air by `1^@C`. Find this volume of air. The specific heat of air is approzimately `1 kJ//kg^@C`. Take the density of air to be `1.3 kg//m^3`.
Answer» The mass of `1m^3` of water is specified by its density, `m=rhoV=(1.00xx10^3 kg//m^(3)))(1m^3)=1xx10^3kg` When `1m^3` of water cools by `1^@C`, it releases energy `Q_c=mcDeltaT=(1xx10^3kg)4186 J//kg.^@C(-1^@C)=-4xx10^6J` Where the negative sign represents heat output. When `+4xx10^6J` is transferred to the air, raising its temperature by `1^@C`, the volume of the air is given by `Q_c=mcDeltaT=rhoVcDeltaT` `V=(Q_c)/(rhocDeltaT)=(4xx10^6J)/((1.3kg//m^3)(1xx10^3J//kg^@C)(1^@C))=3xx10^3m^2` The volume of the air is a thousand times larger than the volume of the water.

Discussion

154.	Give one example where high specific heat capacity of water is used for cooling purposes.
Answer» It is used as coolant by flowing it in pipes around the heated part of machines.

Discussion

155.	A tap supplies water at `10^@ C` and another tap at `100^@ C`. How much hot water must be taken so that we get `20 kg` of water at `35^@ C`A. `40//9 kg`B. `50//9 kg`C. `20//9 kg`D. `130//9 kg`
Answer» Correct Answer - B From principle of calorimetry , `theta = (m_(1)S_(1)theta_(1)+m_(2)S_(2)theta_(2))/(m_(1)S_(1)+m_(2)S_(2))(Given,m_(1)+m_(2) = 20)`.

Discussion

156.	Explain the following :Wet soil does not get as hot as dry soil in the sun.
Answer» Water has high sp. heat capacity as compared to soil (dry) and absorbs heat from surrounding for longer time and takes longer time to set as compared to dry soil.

Discussion

157.	Explain the following :A wise farmer always waters his fields in the evening, if there is a forecast for frost.
Answer» To save the crops on such cold nights farmers fill their fields with water as water has high sp. heat capacity. So water does not allow the temp, in the surrounding area of plants to fall upto 0°C. Other wise when temp, falls below 0°C water in the fine capillaries of plants will freeze, so the veins will burst due to increase in volume of water on freezing.

Discussion

158.	Explain the following :Water is used as a coolant in motor car radiators.
Answer» When water is circulated in the pipes, it absorbs more amount of heat from surroundings (removes heat) without much rise in its temperature because of its high specific heat capacity.

Discussion

159.	Calculate the amount of heat energy required to raise the temperature of 100 g of copper from 20°C to 70°C. specific heat capacity of copper = 390 J kg-1 K-1.
Answer» Mass of copper m = 100 g = 0.1 kg Change of temperature Δt = (70 − 20)^oC Specific heat of capacity of copper = 390 J kg^-1K^-1 Amount of heat required to raise the temperature of 0.1 kg of copper is Q = m × Δt × c = 0.1 × 50 × 390 = 1950 J

159.

Calculate the amount of heat energy required to raise the temperature of 100 g of copper from 20°C to 70°C. specific heat capacity of copper = 390 J kg-1 K-1.

Answer»

Mass of copper m = 100 g = 0.1 kg

Change of temperature Δt = (70 − 20)^oC

Specific heat of capacity of copper = 390 J kg^-1K^-1

Amount of heat required to raise the temperature of 0.1 kg of copper is

Q = m × Δt × c

= 0.1 × 50 × 390

= 1950 J

Discussion

160.	The S.I. unit of specific heat capacity is:(a) J kg-1 (b) J K-1(c) J kg-1 K-1 (d) kilocal kg-1 ℃-1
Answer» The S.I. unit of specific heat capacity is- J kg^-1 K^-1

160.

The S.I. unit of specific heat capacity is:(a) J kg-1 (b) J K-1(c) J kg-1 K-1 (d) kilocal kg-1 ℃-1

Answer»

The S.I. unit of specific heat capacity is-

J kg^-1 K^-1

Discussion

161.	The S.I. unit of heat capacity is:(a) J kg-1 (b) (b) J K-1 (c) (c) J kg-1 K-1 (d) cal ℃-1
Answer» The S.I. unit of heat capacity is- (b) (b) J K^-1

161.

The S.I. unit of heat capacity is:(a) J kg-1 (b) (b) J K-1 (c) (c) J kg-1 K-1 (d) cal ℃-1

Answer»

The S.I. unit of heat capacity is-

(b) (b) J K^-1

Discussion

162.	A heater of power P watt raises the temperature of m kg of a liquid by Δt K in time t s. Express the specific heat capacity of liquid in terms of above data.
Answer» The specific heat capacity of liquid in terms of the above data is: c = P t/m Δ T

162.

A heater of power P watt raises the temperature of m kg of a liquid by Δt K in time t s. Express the specific heat capacity of liquid in terms of above data.

Answer»

The specific heat capacity of liquid in terms of the above data is:

c = P t/m Δ T

Discussion

163.	An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 ℃ to 15.0 ℃ I 100 s. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity of the liquid.
Answer» Power of heater P = 600 W Mass of liquid m = 4.0 kg Change in temperature of liquid = (15 − 10)^oC = 5^oC(or 5 K) Time taken to raise its temperature = 100s Heat energy required to heat the liquid ΔQ = mcΔT And ΔQ = P × t = 600 × 100 = 60000J c = ΔQ/mΔT = 60000/4 x 5 = 3000 Jkg^-1K^-1 Heat capacity = c × m Heat capacity = 4 × 3000 = 1.2 × 10⁴ J/K

163.

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 ℃ to 15.0 ℃ I 100 s. Calculate: (i) the heat capacity of 4.0 kg of liquid, (ii) the specific heat capacity of the liquid.

Answer»

Power of heater P = 600 W

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 − 10)^oC = 5^oC(or 5 K)

Time taken to raise its temperature = 100s

Heat energy required to heat the liquid

ΔQ = mcΔT

And

ΔQ = P × t = 600 × 100 = 60000J

c = ΔQ/mΔT = 60000/4 x 5 = 3000 Jkg^-1K^-1

Heat capacity = c × m

Heat capacity = 4 × 3000 = 1.2 × 10⁴ J/K

Discussion

164.	Differentiate between heat capacity and specific heat capacity.
Answer» Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1 ℃whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1 ℃. Heat capacity of a substance depends upon the material and mass of the body. Specific heatcapacity of a substance does not depend on the mass of the body. S.I. unit of heat capacity is JK^-1 and S.I. unit of specific heat capacity is Jkg^-1 K^-1.

164.

Differentiate between heat capacity and specific heat capacity.

Answer»

Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1 ℃whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1 ℃.

Heat capacity of a substance depends upon the material and mass of the body. Specific heatcapacity of a substance does not depend on the mass of the body.

S.I. unit of heat capacity is JK^-1 and S.I. unit of specific heat capacity is Jkg^-1 K^-1.

Discussion

165.	Define the term ‘Heat capacity’ and state its S.I. unit.
Answer» Heat capacity : The amount of heat energy required to raise the temperature of a given mass of a substance through 1k (or 1°C) is call its heat capacity. S.I. unit of heat capacity is JK^-1.

165.

Define the term ‘Heat capacity’ and state its S.I. unit.

Answer»

Heat capacity : The amount of heat energy required to raise the temperature of a given mass of a substance through 1k (or 1°C) is call its heat capacity.

S.I. unit of heat capacity is JK^-1.

Discussion

166.

Differentiate between heat and temperature.

Answer»

Heat	Temperature
1. Heat is the energy of transit. 2. Its S.I. unit is Joule. 3.It is the measured by the principle of Calorimetry. 4. It is an addictive quantity.	1.Temperature is the fundamental quantity which determines the direction of flow of heat. 2. Its S.I. unit is Kelvin. 3. It is measured by Thermometer. 4. It is not an addictive quantity.

Discussion

167.	Define the term heat capacity and state its S.I. unit.
Answer» The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1^oC (or 1K). S.I. unit is joule per kilogram per kelvin (Jkg^-1K^-1).

167.

Define the term heat capacity and state its S.I. unit.

Answer»

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1^oC (or 1K).

S.I. unit is joule per kilogram per kelvin (Jkg^-1K^-1).

Discussion

168.	Define calorimetry
Answer» The measurement of the quantity of heat is called calorimetry.

Discussion

169.	Define Calorimetry.
Answer» Calorimetery : The measurement of the quantity of heat is called Callorimetery.

Discussion

170.

Differentiate between heat capacity and specific heat capacity.

Answer»

Heat capacity	Sp. Heat capacity
(i) S. I. unit is K^-1 (ii) Heat capacity = mass of sub x sp. heat capacity (iii) It changes with change in mass (iv) It is the amount of heat reqd to raise the temp. of a given mass of substance through 1k or 1°C	S.I. unit JKg^-1k^-1 Sp. heat capacity = (heat capacity)/mass(m) It is constant quantity It is the amount of heat energy reqd. to raise the temp. of a unit mass of substance through 1k or 1°C

Discussion

171.	A cooking vessel on a slow burner contains 5 kg of water and an unknown mass of ice in equilibrium at `0^@C` at time `t=0`. The temperature of the mixture is measured at various times and the result is plotted as shown in Fig. During the first 50 min the mixture remains at `0^@C`. From 50 min to 60 min, the temperature increases to `2^@C` Neglecting the heat capacity of the vessel, the initial mass of the ice isA. `(10)/(7)kg`B. `(5)/(7)kg`C. `(5)/(4)kg`D. `(5)/(8)kg`
Answer» Correct Answer - B Let m be the mass of ice. Rate of heat given by the burner is constant. In the first 50 min `(dQ)/(dt)=(mL)/(t_1)=(mkgxx(80xx4.2xx10^3)J//kg)/((50min))` .(i) From 50 min to 60 min `(dQ)/(dt)=((m+5)S_(H_2O)Deltatheta)/(t_2)` `=((m+5)kg(4.2xx10^3)J//kg xx2^@C)/(10min)` ..(ii) From Eqs. (i) and (ii) `(80m)/(50)=(2(m+5))/(10)` `7m=5impliesm=(5)/(7)kg=0.7kg`

Discussion

172.	An equal quantity of heat is supplied to two substances A and B. The substance A shows a greater rise in temperature. What can you say about the heat capacity of A as compared to that of B?
Answer» Heat absorbed by a substance is given by H = msθ H = Heat capacity × rise of temperature. Since, H is same for both A and B, it is a clear that heat capacity is inversely proportional to the rise of temperature. Since, the rise of temperature A is more its heat capacity must be less. ∴ Heat capacity of A is less than that of B.

172.

An equal quantity of heat is supplied to two substances A and B. The substance A shows a greater rise in temperature. What can you say about the heat capacity of A as compared to that of B?

Answer»

Heat absorbed by a substance is given by

H = msθ

H = Heat capacity × rise of temperature.

Since, H is same for both A and B, it is a clear that heat capacity is inversely proportional to the rise of temperature.

Since, the rise of temperature A is more its heat capacity must be less.

∴ Heat capacity of A is less than that of B.

Discussion

173.	Explain why some rubber-like substances contract with rising temperature.
Answer» Atoms in a solid vibrate longitudinally as well as transversely. The average separation of the planes of atoms increases due to their longitudinal vibrations and decreases due to their transverse vibrations with rise in temperature. Expansion depends on the dominance of one type of vibration over the other. In some crystals, the longitudinal vibration dominate over the transverse vibration. Such substances expand on heating. In some other crystals like rubber (sonsisting of long interwined and cross-linked chains of atoms in roughly randon orientation), transverse vibrations dominate over longitudinal vibration. Such solids contract on heating.

Discussion

174.	A meteorite has mass of 500 kg and is composed of a metal. The temperature of the meteor is `-20^(@)C` and its speed is 10 km/hr when it is at large distance from a planet. The meteorite crashes into the planet and its entire kinetic energy gets converted into heat. This heat is equally shared between the planet and the meteorite. Assume that the heating of meteorite is uniform and the average specific heat capacity of the metal, for its solid, liquid and vapour phase, is `1200 J kg^(-1) .^(@)C^(-1)`. The latent heat of fusion and vaporization of the metal are `L_(f) = 4 xx 10^(5) J kg^(-1)` and `L_(v) = 1.1 xx 10^(7) J kg^(-1)` respectively. The melting point and boiling points are `380 .^(@)C` and `2380 .^(@)C` respectively. Find the temperature of the meteorite material immediately after the impact. Take: `G = 6.6 xx 10^(11) N m^(2) kg^(-2)`, mass of planet `M = 6 xx 10^(24) kg`, radius of planet R = 6600 km
Answer» Correct Answer - Nearly `36400^(@)C`

Discussion

175.	The ratio of spacific to moler head capacity of a bodyA. is a universal constantB. depends on the mass of the bodyC. depends on the molecular weight of the bodyD. is dimensionless
Answer» Correct Answer - C depends on the molecular weight of the body

Discussion

176.	An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in the pitcher to cool to `28^@C` from its original temperature of `30^@C` Neglect radiation effect. Latent heat of vapourization of water in this range of temperature is 580 cal/g and specific heat of water is `1 kcal//gC^(@)`
Answer» Here water at the surface is evaporated at the cost of the water in the vessel losing heat. Heat lost by the water in the vessel `=(9.5+0.5)xx1000xx(30-20)=10^5`cal Let t be the required time in seconds. Heat gained by the water at the surface `=(txx10^-3)xx540xx10^3` `(L=540(cal)/(g)=540xx10^3cal//kg)` `:. 10^5=540t` or `t=185s=3min5s`

Discussion

177.	An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in the pitcher to cool to `28^@C` from its original temperature of `30^@C` Neglect radiation effect. Latent heat of vapourization of water in this range of temperature is 580 cal/g and specific heat of water is `1 kcal//gC^(@)`A. 38.6 minB. 30.5 minC. 34.5 minD. 41.2 min
Answer» Correct Answer - A As water equivalent of pitcher is 0.5 kg i.e.k pitcher is equivalent to 0.5 kg of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from 30 to `28^@C` is `Q_1=(m+M)cDeltaT` `=(9.5+0.5)kg((1kcal)/(kgC^@))(30-28)^@C` And het extractged from the pitcher through evaporation in t minutes `Q_2=mL=[(dm)/(dt)xxt]L=[(1g)/(min)xxt]580 cal//g` According to given problem `Q_2=Q_1`, i.e., `580xxt=20xx10^3` `t=34.5min`

Discussion

178.	A liquid with coefficient of volume expansion `gamma` is filled in a container of a material having coefficient of linear expansion `alpha` . If the liquid overflows on heating, thenA. `gamma gt 3alpha`B. `gamma lt 3alpha`C. `gamma=3alpha`D. none of these
Answer» Correct Answer - A `V_(l)xxV_(e)` `gamma gt 3alpha`.

Discussion

179.	Indian style of cooling drinking water is to kept it is a pitcher having porous walls water comes to the outer surface very alowly and evaporates .Most of the itself and the water is cooles down .Assums that a pitcher containe `10kg` water and `0.2g` of water comes from the itomsphere decrease by ``5^(@)C` specific beat capicity of water` = 4200J kg^(-1)^(@)C^(-1)` and letent head of vaporization of water `= 2.27 xx 10^(6)J kg^(-1)`
Answer» Energy required to decreases the temp of `10kg` of water to `5^(@)C` `U = 10 xx 4200J//kg^(@)C xx 5^(@)C` `= 210.000 = 21 xx 10^(6)J` `Energy required to for evaporation of water to `0.2kg//sec` `= 2 xx 10^(-4) xx 2.27 xx 10^(6)` = 454` `454J` energy losing system per second `= (21 xx 10^(4))/(454)` `21 xx 10^(4)J`energy losing system during one minute `= (21 xx 10^(4))/(454 xx 60) = 7.7 minute` The time required to decrease temp by `5^(@)C is 7.7 minute`

Discussion

180.	Consider a cylindrical container of cross-section area A length h and having coefficient of linear expansion `alpha_(c)`. The container is filled by liquid of real expansion coefficient `gamma_(L)` up to height `h_(1)`. When temperature of the system is increased by `Deltatheta` then (a). Find out the height, area and volume of cylindrical container and new volume of liquid. (b). Find the height of liquid level when expansion of container is neglected. (c). Find the relation between `gamma_(L)` and `alpha_(c)` for which volume of container above the liquid level (i) increases (ii). decreases (iii). remains constant. (d). On the surface of a cylindrical container a scale is attached for the measurement of level of liquid of liquid filled inside it. If we increase the temperature of the temperature of the system by `Deltatheta`, then (i). Find height of liquid level as shown by the scale on the vessel. Neglect expansion of liquid. (ii). Find the height of liquid level as shown by the scale on the vessel. Neglect expansion of container.
Answer» On increasing the temperature, the height area of cross section and volume of the cylinder will increase. (a). New height `=h_(f)=h{1+alphaDeltatheta}` New area of cross section `A_(f)=A{1+2alpha_(c)Deltatheta}` New volume of container `V_(f)=Ah{1+3alpha(c)Deltatheta}` New volume of liquid `V_(omega)=V_(0)(1+gamma_(L)Deltatheta)` `V_(omega)=Ah_(1)(1+gamma_(L)Deltatheta)` (b). The height of liquid level when expansion of container is neglected `h_(f)=(V_(omega))/(A)=(Ah_(1)(1+Y_(L)Deltatheta))/(A)` `impliesh_(f)=h_(1){1+gamma_(L)Deltatheta}` (c). The initial volume of container above the liquid. `DeltaV_(1)=Ah-Ah_(1)` final volume of container above the liquid `DeltaV_(2)=Ah(1+3alpha_(c)Deltatheta)-Ah_(1)(1+gamma_(L)Deltatheta)` If volume of above container increases `DeltaV_(2)gtDeltaV_(1)` `[Ah(1+3alpha_(c)Deltatheta)-Ah_(1)(1+gamma_(L)Deltatheta)]gt[Ah-Ah_(1)]` Which gives `3halpha_(c)gth_(1)gamma_(L)` Similarly, we can prove if the volume above container decreases, then `3halpha_(c)lth_(1)gamma_(L)` and for no change in volume 3 h `alpha_(c)=h_(1)gamma_(L)` (d). The area of container will increase, the area of container at this temperature will be `A_(f)=A{1+2alpha_(c)Deltatheta}` As liquid does not expand the volume of the liquid wll be as initial volume `=Ah_(1)` Hence height of the liquid column will be `h_(1)^(`)=`(Ah_(1))/(A_(f))=(Ah_(1))/(A(1+2alpha_(c)Deltatheta))impliesh_(1)^(`)=`H_(1)(1-2alpha_(c)Deltatheta)` . In this case we are neglecting the expansion of container. The volume of liquid at this temperature ` ` `V_(omega^(`))`=`Ah_(1)(1+gamma_(L)Deltatheta)` ` ` Hence height of the liquid in container is `h_(f)`=`(V_(omega))/(A)`=`(Ah_(1)(1+gamma_(L)Deltatheta))/(A)` implies `h_(f)`=`h_(1)(1+gamma_(L)Deltatheta))/(A)`implies `h_(f)`=`h_(1)(1+gamma_(L)` `Deltatheta)`

Discussion

181.	The coefficient of linear expansion of crystal in one direction is `alpha_(1)` and that in every direction perpendicular to it is `alpha_(2)`. The coefficient of cubical expansion isA. `alpha_1+alpha_2`B. `2alpha_1+alpha_2`C. `alpha_1+2alpha_2`D. none of these
Answer» Correct Answer - C `V=V_0(1+gammaDeltatheta)` or `L^3=L_0(1+alpha_1Deltatheta)L_0^2(1+alpha_2Deltatheta)^2` `impliesL^3=L_0^3(1+alpha_1Deltatheta)(1+alpha_2Deltatheta)^2` since `L_0^3=V_0`, hence `1+gammaDeltatheta=(1+alpha_1Deltatheta)(1+alpha_2Deltatheta)^2` `cong(1+alpha_1Deltatheta)(1+2alpha_2Deltatheta)` `cong 1+alpha_1Deltatheta+2alpha_2Deltatheta` `:. gamma_(1) + 2 alpha_(2)`.

Discussion

182.	The coefficient of linear expansion of an in homogeneous rod change linearly from `alpha_(1)` to `alpha_(2)` from one end to the other end of the rod. The effective coefficient of linear expansion of rod isA. `alpha_1+alpha_2`B. `(alpha_1+alpha_2)/(2)`C. `sqrt(alpha_1alpha_2)`D. `alpha_1-alpha_2`
Answer» Correct Answer - B The effective value of `alpha` at a distance x from the left end is `alpha_x=alpha_1+((alpha_2+alpha_1)/(L))x` `DeltaL=int_0^Lalpha_xdxDeltat` `L=((alpha_1+alpha_2)/(2))LDeltaT` `alpha_("eff")=(alpha_1+alpha_2)/(2)`

Discussion

183.	100 g of ice at `-40^(@)C` is supplied heat using a heater. The heater is switched on at time t = 0 and its power increases linearly for first 60 second and thereafter it becomes constant as shown in the graph. Heater is kept on for 5 minutes. The specific heat capacity for ice and water are known to be `2.1 (J)/(g^(@)C)` and `4.2 (J)/(g^(@)C)` respectively. The specific latent heat for fusion of ice is `336 J//g`. The temperature of the ice sample kept on increasing till time `t_(1)` and then remained constant in the interval `t_(1) lt t lt t_(2)`. (i) Find `t_(1)` and `t_(2)` (ii) Find final temperature of the sample when the heater is switched off.
Answer» Correct Answer - (i) `t_(1) = 65s; t_(2) = 265s`, (ii) `14^(@)C`

Discussion

184.	A piece of ice (heat capacity `=2100 J kg^(-1) .^(@)C^(-1)` and latent heat `=3.36xx10^(5) J kg^(-1)`) of mass m grams is at `-5 .^(@)C` at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice . Water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m in gram is
Answer» Here, heat given is used to increase the temperature of the ice to `0^@ C` and to melt `1 gm` of ice. Given `m` is mass of ice in `gm`. :. `420 = (m xx 2100 xx 5 + 1 xx 3.36 xx 10^(5)) xx 10^(-3)` `rArr m = 8 gm`.

Discussion

185.	The amount of heat supplied to decrease the volume of an ice water mixture by `1 cm^(3)` without any change in temperature, is equal to: `(rho_("ice") = 0.9, rho_("water") = 80 cal//gm)`A. 360 calB. 500calC. 72 calD. 720 cal
Answer» Correct Answer - C `Q = m xx L_(ice) = rho_(ice)(V) L_(ice)`.

Discussion

186.	An ice ball has a metal piece embedded into it. The temperature of the ball is `-theta^(@)C` and it contains mass M of ice. When placed in a large tub containing water at `0^(@)C`, it sinks. Assume that the water in immediate contact with the ice ball freezes and thereby size of the ball grows. What is the maximum possible mass of the metal piece so that the ball can eventually begin to float. Densities of ice, water and metal are `sigma, rho` and d respectively. Specific heat capacity of ice is s and its specific latent heat is L. Neglect the heat capacity of the metal piece.
Answer» Correct Answer - `(d)/(sigma)((rho-sigma))/((d-rho)) [1+(s theta)/(L)]`

Discussion

187.	A well insulated container has a mixture of ice and water, at `0^(@)C`. The mixture is supplied heat at a constant rate of 420 watt by switching on an electric heater at time t = 0. The temperature of the mixture was recorded at time t = 150s, 273s and 378s and the readings were `0^(@)C`, `10^(@)C` and `20^(@)C` respectively. Calculate the mass of water and ice in the mixture. Specific heat of water `= 4.2 J g^(-1) .^(@)C^(-1)`, Specific latent heat of fusion of ice `= 336 J g^(-1)`. Assume that the mixture is stirred slowly to maintain a uniform temperature of its content.
Answer» Correct Answer - `m_(omega) = 840 g; m_("ice") = 210 g`

Discussion

188.	A well insulated box has two compartments A and B with a conducting wall between them. 100 g of ice at `0^(@)C` is kept in compartment A and 100 g of water at `100^(@)C` is kept in B at time t = 0. The temperature of the two parts A and B is monitored and a graph is plotted for temperatures `T_(A)` and `T_(B)` versus time (t) [Fig. (b)]. Assume that temperature inside each compartment remains uniform. (a) Is it correct to assert that the conducting wall conducts heat at a uniform rate, irrespective of the temperature difference between A and B? (b) Find the value of time `t_(1)` and temperature `T_(0)` shown in the graph, if it is known that `t_(0) = 200 s`. Specific heat of ice `= 0.5 cal g^(-1) .^(@)C^(-1)` Specific heat of water `= 1.0 cal g^(-1) .^(@)C^(-1)` Latent heat of fusion of ice `= 80 cal g^(-1)`
Answer» Correct Answer - (a) yes , (b) `T_(0) = 13.3^(@)C, t_(1) = 217 s`

Discussion

189.	Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water, both at 0° C?
Answer» Every gram of ice surrounding the soft drink extracts out 336J of heat energy from it and the temperature of surrounding the soft drink remains at 0°C. However, in case of cold water, it will extract out only 4.2J of heat energy per gram. Furthermore, the temperature of surrounding water starts rising. Thus, soft - drink bottle cools better in case of ice.

189.

Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water, both at 0° C?

Answer»

Every gram of ice surrounding the soft drink extracts out 336J of heat energy from it and the temperature of surrounding the soft drink remains at 0°C. However, in case of cold water, it will extract out only 4.2J of heat energy per gram.

Furthermore, the temperature of surrounding water starts rising.

Thus, soft - drink bottle cools better in case of ice.

Discussion

190.	A certain amount of heat Q will warm 1g of material X by 3°C and 1g of material Y by 4°C. Which material has a higher specific heat capacity.
Answer» Material X has higher specific heat capacity compared to material Y. It is because for the mass amount of heat its temperature rises less than Y.

190.

A certain amount of heat Q will warm 1g of material X by 3°C and 1g of material Y by 4°C. Which material has a higher specific heat capacity.

Answer»

Material X has higher specific heat capacity compared to material Y.

It is because for the mass amount of heat its temperature rises less than Y.

Discussion

191.	Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) isA. `27.33 g`B. `24.8 g`C. `2.8 g`D. `30 g`
Answer» Correct Answer - B `m_(steam)L_(v) +m_(steam) S_(w) (100^@ -90^@)` =`m_(w)S_(w)(90^@-20^@)` `m_(steam)` = mass f steam converted into water :. Mass of water `= 22 g + m_(steam)`.

Discussion

192.	A `10 kW` drilling machine is used to drill a bore in a small aluminium block of mass `8.0 kg`. How much is the rise in temperature of the block in `2.5` minutes, assuming `50%` of power is used up in heating the machine itself or lost to the surrounding? Specific heat of aluminium `= 0.91 J//g^(@)C`.A. `103^@C`B. `130^@C`C. `105^@C`D. `30^@C`
Answer» Correct Answer - A Total work done by drill machine in `2.5xx60s` `=(10xx10^3)(2.5xx60)=15xx10^5J` Energy lost `=50%` of `15xx10^5J=7.5xx10^5J` Energy taken by its surroundings, i.e., aluminium block `DeltaQ=mcDeltat=8xx10^3xx0.91xxDeltaTJ` Energy given `=`Energy taken `7.5xx10^5=8xx10^3xx0.91xxDeltaT` `impliesT=103^@C`

Discussion

193.	Three rods of same dimensions are arranged as shown in Fig. They have thermal conductivities `K_1`,`K_2` and `K_3`. The points P and Q are maintained at different temeperature for the heat to flow at the same rate along PRQ and PQ. Whi of the following options correct?A. `K_3=(1)/(2)(K_1+K_2)`B. `K_3=K_1+K_2`C. `K_3=(K_1K_2)/(K_1+K_2)`D. `K_3=2(K_1+K_2)`
Answer» Correct Answer - C Rate of flow of heat along PQ `((dQ)/(dt))_(PQ)=(K_3ADeltatheta)/(l)` .(i) Rate of flow of heat along PRQ `((dQ)/(dt))_(ORQ)=(K_SaDeltatheta)/(2l)` Effective conductivity for series combination of two rods of same length `K_S=(2K_1K_2)/(K_1+K_2)` So `((dQ)/(dt))_(PRQ)=(2K_1K_2)/(K_1+K_2)(ADeltatheta)/(2l)=(K_1K_2)/(K_1+K_2)(ADeltatheta)/(l)` .(ii) Equating Eq. (i) and (ii) `K_3=(K_1K_2)/(K_1+K_2)`

Discussion

194.	Two blocks of masses `10kg and 20kg` moving at speeds of `10ms^(-1) and 20ms^(-1)` respectively in opposite direction approach each other and colide .If the collision is completely inelastic , find the thermal energy developed in the process
Answer» Correct Answer - C `m_(1) = 10kg,V_(1) = 10m//s` `m_(2) = 20kg,V_(2) = 20m//s` Here `m_(2)V_(2) - m_(1)V_(1) = (m_(1) + m_(2))V` `rArr 20 xx 20 - 10 xx 10 = (10 + 20)V` `rArr 400 - 100 = 30V` rArr 300 = 30V` rArr V =10m//s` Initial kinetic energy `= (1)/(2) m_(1)u_(1)^(2)+ (1)/(2)m_(2)u_(2)^(2)` `= (1)/(2) xx 10 xx (10)^(2) + (1)/(2) xx 20 xx (20)^(2)` ` = 500 + 4000 = 4500` Final kinetic energy `= (1)/(2) (m_(1) +m_(2))V^(2)` `= (1)/(2) (10+20)(10)^(2)` `= ((30)/(2)) xx 100 = 1500` THe total change on K.E.`= 4500 - 1500= 3000J` The thermal enrgy devloped in the process `= 3000J`

Discussion

195.	Define specific latent heat of fusion of ice. State its magnitude in calories and joules.
Answer» Specific latent heat of fusion of ice : “Is the heat energy required to convert unit mass of ice to water without the change in temp, (or ice at 0°C to water at 0°C).” Specific latent heat of fusion of ice = 80cal g^-1 or 336000Jkg^-1

Discussion

196.	Why should bits of ice to wiped dry before adding them to the calorimeter during the determination of specific latent heat of fusion of ice?
Answer» If bits of ice are not wiped dry waterdrops are already in liquid state will absorb less heat and result will not be correct.

Discussion

197.	What do you understand by the term specific latent heat of fusion? State its C.GS. and S.I. unit.
Answer» Specific latent heat of fusion : “The heat energy required to convert unit mass of the substance from solid to liquid state without change in temperature.” Units : C.G.S. → Cal g^-1 S.I. Jkg^-1

197.

What do you understand by the term specific latent heat of fusion? State its C.GS. and S.I. unit.

Answer»

Specific latent heat of fusion : “The heat energy required to convert unit mass of the substance from solid to liquid state without change in temperature.”

Units :

C.G.S. → Cal g^-1

S.I. Jkg^-1

Discussion

198.	The specific latent heat of fusion of ice is 336 J g-1. Explain the meaning of its statement.
Answer» It means 1 g of ice at 0^oC absorbs 360 J of heat energy to convert into water at 0^oC.

Discussion

199.	The specific latent heat of fusion of water is:(a) 80 cal g-1 (b) 2260 J g-1(c) 80 J g-1 (d) 336 J kg-1
Answer» (a) 80 cal g^-1 .

Discussion

200.

The S.I. unit of specific latent heat is:(a) cal g-1 (b) cal g-1K-1(c) J kg-1 (d) J kg-1 K-1

Answer»

The S.I. unit of specific latent heat is-

(c) J kg^-1

Discussion

Explore topic-wise InterviewSolutions in .

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A certain amount of heat Q will warm 1g of material X by 3°C and 1g of material Y by 4°C. Which material has a higher specific heat capacity.

Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) isA. `27.33 g`B. `24.8 g`C. `2.8 g`D. `30 g`

Two blocks of masses `10kg and 20kg` moving at speeds of `10ms^(-1) and 20ms^(-1)` respectively in opposite direction approach each other and colide .If the collision is completely inelastic , find the thermal energy developed in the process

Define specific latent heat of fusion of ice. State its magnitude in calories and joules.

Why should bits of ice to wiped dry before adding them to the calorimeter during the determination of specific latent heat of fusion of ice?

What do you understand by the term specific latent heat of fusion? State its C.GS. and S.I. unit.

The specific latent heat of fusion of ice is 336 J g-1. Explain the meaning of its statement.

The specific latent heat of fusion of water is:(a) 80 cal g-1 (b) 2260 J g-1(c) 80 J g-1 (d) 336 J kg-1

The S.I. unit of specific latent heat is:(a) cal g-1 (b) cal g-1K-1(c) J kg-1 (d) J kg-1 K-1