ICSE Physics Class X (HEAT) Important Questions (Solved)

HEAT

Important Questions (Solved)


Question .1. (i) Define specific latent heat of vaporisation.

(ii) Draw a labelled diagram of the apparatus you would use to determine the specific latent heat of vaporisation of steam by the method of mixture.

(iii) State two precautions you would take while performing the experiment with the apparatus.

(iv) In an experiment to determine the specific latent heat of vaporisation of steam L, the following measurements were taken :

Mass of calorimeter + stirrer = x kg

Mass of water = y kg

Initial temperature of water = t1ºC

Final temperature of mixture = t2ºC

Given : Specific heat capacity of calorimeter and water are S1 and S2 respectively. Express L in terms of above data. Mass of condensity (steam condensed) = m kg.

Solution:

(i) The specific latent heat of vaporisation of a liquid is defined as the heat required to convert one kilogram of the liquid into vapour at its boiling point without any change in its temperature.

(ii) For diagram contact a T.B.

(iii) Precautions : (1) The calorimeter must be separated from the part of the apparatus to produce steam by thermally opaque screen and

(2) Steam trap must be used to minimize the chances of condensing of steam getting into the calorimeter.

(iv) Heat lost by steam to condense = m × L

Heat lost by water to lower its temperature to t2 = m × S2 (100 – t2)

Total heat lost = m × L + m × S2 (100 – t2) ———————– (1)

Heat gained by water = y × S2 (t2 – t1)

Heat gained by calorimeter = x × S1 (t2– t1)

Total heat gained = y × S2(t2 – t1) + x × S1 (t2– t1) ——————- (2)

By the principle of Calorimetry, Heat lost = Heat gained, we get

mL + mS2 (100 –t2) = yS2 (t2 – t1) + xS1 (t2 – t1)

Or, L = {(xS1 + yS2)(t2 – t1) – mS2 (100 – t2)}/ m

Question .2. Which material is the calorimeter commonly made of?

Solution:

Calorimeter is made of copper.

Question .3. Give two reasons as to why copper is preferred over other metals for making calorimeter. 

Solution:

(i) Being good conductor of heat calorimeter made of copper acquires the temperature of its contents.

(ii) The thermal capacity of calorimeter is low as specific heat capacity of copper is low.

Question .4. Define the term specific heat capacity of a substance. Give its SI unit.

Solution:

Specific latent heat is defined as the amount of heat required to raise the temperature of one kilogram of a substance through 1 ºC. The SI unit of specific heat capacity is Jkg-1 ºC.

Question .5. Define specific latent heat of fusion of ice. Give its SI unit.

Solution:

The specific latent heat of fusion of ice is defined as the heat required to melt one kilogram of ice at its melting point, without any change of temperature. SI unit of specific latent heat of fusion is Jkg-1.

Question .6. Explain briefly: Why hot water bottles are very efficient for fomentation?  

Solution:

Hot water bottles are very efficient for fomentation, because water can store in it a large amount of heat energy at fairly low temperature, owing to its high specific heat capacity.

Question .7. What is the effect of increase in pressure on the melting point of ice and boiling point of water?

Solution:

(i) Melting point of ice decreases when pressure is increased.

(ii) Boiling point of water increases when pressure is increased.

Question .8. Explain why bottled soft drinks are more effectively cooled by cubes of ice than by ice water.

Solution:

Ice needs heat equal to its specific latent heat of fusion (336 ×103 Jkg-1) to melt and form ice-cold water at 0 ºC. It means each kg of ice can withdraw 336 × 103 J more heat from the soft drink bottle than ice cold water at 0 ºC. Hence soft drinks are more effectively cooled by ice cubes.

Question .9. Explain why the surroundings become pleasantly warm when water in a lake starts freezing in cold countries.

Solution:

When water freezes it releases heat equal to its latent heat of fusion into the surroundings, which makes the surrounding pleasantly warm.

Question .10. If, in a central heating system, steam enters a radiator pipe at 100 ºC and water leaves the radiator pipe at 100 ºC, can this radiator pipe heat a room? Give an explanation for your answer.

Solution:

Yes, because steam at 100 ºC releases heat equal to its latent heat of vaporization (22,60,000 Jkg-1) to condense into water at 100 ºC.

Question .11. It takes a much longer time to boil off (change to steam) a certain quantity of water, rather than to bring it to its boiling point from room temperature, say 25 ºC. Explain the reason for this.

Solution:

This is due to the fact that latent heat of vaporisation of water is 2268000 Lkg-1 in comparison to the specific heat of water which is 4200 J kg-1 ºC-1. Thus the heat required to boil off a certain quantity of water (mL) is more than to raise the temperature of water to its boiling point (mcθ).

Question .12. Explain why one feels ice cream at 0 ºC colder than water at 0 ºC.

Solution:

Ice cream needs latent heat of fusion to melt. Latent heat of fusion of water is more than the specific heat. Hence ice cream is colder than water both at 0 ºC.

Question .13. Give one example of each where high specific heat capacity of water is used (i) in cooling and (ii) as heat reservoir.

Or,

Explain why water is used in hot water bottles for fomentation and also as a universal coolant.

Solution:

(i) Water is used to cool the radiators of engines of cars.

(ii) As a heat reservoir water is used in hot water bottles for fomentation or in hot water pipes for heating rooms in cold countries.

Question .14. Does land cool at a slower or faster rate than water? Give one reason for your answer.

Solution:

The land cool at a faster rate than water, as specific heat capacity of water is more than that of land.

Question .15. Explain why steam pipes warm a building more effectively than hot water pipes in cold countries.

Solution:

As heat content of 1 kg of steam (at 100 ºC) is more than that of 1 kg of boiling water (also at 100 ºC) by nearly 2260 kJ. Each kg of steam can give out nearly 2260 kJ of heat more than each kg of boiling hot water. Thus steam is more effective than boiling hot water for warming a building.

Question .16. In winter, the weather forecast for a certain day was ‘severe frost’. A wise farmer watered his fields the night before to prevent frost damage to his crops. Why did he water his fields?     

Solution:

Due to severe frost, the water present in the field would freeze and release heat and this heat prevent damage of crops (water present inside the crop will not freeze).

Question .17. State the principle of Calorimetry.

Or,

What is the principle of Calorimetry?

Solution:

When two objects exchange heat then the amount of heat lost by one is equal to the amount of heat gained by the other. In short we say :

Heat gained = Heat lost.

Question .18. Which of the two , 1 g of ice at 0 ºC or 1 g of water at 0 ºC contains more heat? Give reason for your answer.

Solution:

1 g of water at 0 ºC contains more heat than 1 g of ice at 0 ºC. When 1 g of water is converted into ice, releases 80 calories of heat. Hence 1 g of water at 0 ºC contains 336 J of heat more than 1 g of ice at 0 ºC.

Question .19. Why are burns caused by steam more severe than those caused by boiling water at the same temperature?

Solution:

Steam loses more heat than boiling water in cooling. Every 1 g of steam loses 2260 J of heat in changing to water at 100 ºC. Thus steam cause more severe burns than water at 100 ºC

Question .20. Why does the heat supplied to a substance during its change of state not cause any rise in its temperature?

Solution:

The heat supplied during change of state does not appear on thermometer as it is used up in bringing about a change of state. There is a change in the distance between the molecules.


PROBLEMS FOR PRACTICE


Numerical – 1. Some hot water was added to three times its mass of cold water at 10 ºC and the resulting temperature was found to be 20 ºC. What was the temperature of the hot water?

Solution:

Let mass of hot water = m

Then mass of cold water = 3 m

Temperature of mixture = 20 ºC

Let temperature of hot water = t ºC

Heat lost by hot water in cooling from t ºC to 20 ºC = m × c × (t – 20)

Heat gained by cold water = 3 m × c × (20 – 10) = 30 mc

By the principle of Calorimetry : heat lost = heat gained , we have

mc (t – 20 ) = 30 mc => t – 20 = 30 = > t = 50 ºC.

Thus temperature of hot water = 50 ºC.

Numerical – 2. A piece of iron of mass 2.0 kg has a thermal capacity of 966J/ºC.

(i) How much heat is needed to warm it by 15 ºC?

(ii) What is its specific heat capacity in S.I. units.

Solution:

(i) Heat needed = Thermal capacity × rise in temperature

= 966 × 15 = 14490 J.

(ii) Specific heat capacity = Thermal capacity/mass = 966/2 = 483 J/kg ºC.

Numerical – 3. Calculate the amount of heat released when 5.0 g of water at 20 ºC is changed into ice at 0 ºC. Specific heat capacity of water = 4.2 J/g ºC, specific latent heat of fusion of ice = 336 J/g.

Solution:

Amount of heat released = m c ΔT + m L

= 5 × 4.2 × (20 – 0) + 5 × 336

= 420 + 1680= 2100 J

Numerical – 4. m1 g of steam at 100 ºC, is condensed to form water at 100 ºC. If there is no heat loss to the surroundings, the heat released in this process is used to convert m2 g of ice at 0 ºC into water at 0 ºC. Find :

(i) The heat lost by steam in terms of m1.

(ii) The heat gained by ice in terms of m2.

(iii) Form a heat equation and find the ratio of m2 : m1 from it. (Specific latent heat of vaporization of steam = 2268 kJkg-1; Specific latent heat of fusion of ice = 336 kJkg-1; Specific heat capacity of water = 4200 Jkg-1ºC-1).

Solution:

(i) Heat lost by steam = m1 L = m1 × 2268kJkg-1.

(ii) Heat gained by ice = m2 L = m2 × 336 kJkg-1.

(iii) Heat gained by ice = Heat lost by steam

Or, m2 × 336 = m1 × 2268

Or, m2:m1 = 2268/336 = 27:4.

Numerical – 5. Calculate the heat energy that will be released when 5.0 kg of steam at 100 ºC condenses to form water at 100 ºC. Express your answer in SI unit. (Specific latent heat of vaporization of steam is 2268 kJkg-1)

Solution:

Heat energy released by steam = m L = 5.0 × 2268 = 11340 kJ

= 1134 × 104 J.

Numerical – 6. A thermos flask of negligible heat capacity contains 100 g of ice and 30 g of water. Calculate :

(i) the mass of steam at 100 ºC needed to condense in the flask so as to just melt the ice;

(ii) the amount of water in the flask after condensation. (Specific latent heat of vaporization of steam = 2260 Jg-1, Specific latent heat of fusion of ice = 336 Jg-1, Specific heat capacity of water = 4.2 Jg-1 ºC)

(iii) Is it possible to condense the water formed, back to ice by adding ice at 0 ºC? Explain giving a suitable reason to justify your answer.

Solution:

(i) Mass of ice = 100 g

Mass of water = 30 g

The heat required to just melt 100 g ice at 0 ºC = m L

= 100 × 336 = 33600 J.

Let m g of steam is condensed to release these heat. Heat released by m g of steam at 100 ºC to be condensed into water at 100 ºC =

L = m × 2260 = 2260m J.

Heat released by m g of water at 100 ºC in cooling to water at 0 ºC

= m c θ = m × 4.2 × (100 – 0) = 420 m J.

Total heat released by m g of steam = 2260 m + 420 m = 2680 m.

By principle of Calorimetry , heat lost = heat gained

2680 m = 33600 => m = 33600/2680 = 12.5 g.

(ii) Total mass of water in the flask after condensation

= 100 g + 30 g + 12.5 g = 142.5 g.

(iii) No, because water at 0 ºC cannot release 336 J per g of heat to condense it back to ice.

Numerical – 7. 10125 J of heat energy boils off 4.5 g of water at 100 ºC to steam at 100 ºC. Find the specific latent heat of steam.

Solution:

Heat energy used to boil water at 100 ºC = m L

Or, 4.5 × L = 10125 => L = 10125/4.5 = 2250 Jg-1.

Numerical – 8.  In a laboratory experiment for finding the specific latent heat of ice, 100 g of water at 30 ºC was taken in a calorimeter made of copper and mass 10 g. When 10 g of ice at 0 ºC was added to the mixture and kept within the liquid till the ice melted completely, the final temperature of the mixture was found to be 20 ºC:

(i) what is the total quantity of water in the calorimeter at 20 ºC

(ii) Specific heat capacities of water and copper being 4.2 J g ºC-1 and 0.4 J g º respectively, what quantity of heat would each release in cooling down to 20 ºC from the initial stage?

(iii) Write an expression for the heat gained by ice on melting.

(iv) Calculate the value of the latent heat of fusion of ice from the data discussed above.

Solution:

(i) Total mass of water in the calorimeter at 20 ºC

= mass of water taken + mass of water formed after melting the ice

= 100 g + 10 g = 110 g.

(ii) Heat released by water = m c θ = 100 × 4.2 × (30 – 20) = 4200 J

Heat released by calorimeter = m c θ = 10 × 0.4 × 10 = 40 J

(iii) Heat gained by ice to melt = total heat lost by water and calorimeter

= 4200 J + 40 J = 4240 J

(iv) Let L be the specific latent heat of fusion of ice, then

10 × L + 10 × 4.2 × (20 – 0) = 4240

Or, 10L + 840 = 4240 => 10L = 4240 – 840 = 3400

Or, L = 340 Jg-1.

Numerical – 9. A vessel of negligible heat capacity contains 60 g of ice in it at 0 ºC. 8 g of steam at 100 ºC is passed into the ice to melt it. Find the final temperature of the contents of the vessel (specific latent heat of fusion of ice = 336 Jg-1, specific heat capacity of water = 4.2 Jg-1 ºC-1).

Solution:

Let final temperature of the contents of the vessel be t ºC.

Heat given out by 8 g of steam at 100 ºC in condensing into water at 100 ºC

= m L = 8 × 2268 = 18144 J

Heat given out by this 8 g of water at 100 ºC in cooling down to t ºC

= m c θ = 8 × 4.2 × (100 – t) = 3360 – 33.6t

Total heat given out = 18144 + 3360 – 33.6t = (21504 – 33.6t) J

Heat required by 40 g of ice at 0 ºC to melt into water at 0 ºC

= m L = 40 × 336 = 13440 J

Heat required by this 40 g of water at 0 ºC to be heated up to the final temperature of t ºC = m c θ = 40 × 4.2 × (t – 0) = 168t

Total heat taken = (13440 + 168t) J

By principle of Calorimetry , heat gained = heat lost

2344 + 168t = 21504 – 33.6t => t = 40 ºC

The final temperature of the contents of vessel = 40 ºC.

Numerical – 10. In a laboratory experiment to measure specific heat capacity of copper, 0.02 kg of water at 70 ºC was poured into a copper calorimeter with a stirrer of mass 0.16 kg initially at 15 ºC. After stirring the final temperature reached to 45 ºC. Specific heat of water is taken as 4200 Jkg-1ºC-1.

(i) What is the quantity of heat released per kg of water per 1 ºC fall in temperature?

(ii) Calculate the heat energy released by water in the experiment in cooling from 70 ºC to 45 ºC.

(iii) Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 15 ºC to 45 ºC, calculate the specific heat capacity of copper.

Solution:

(i) The heat energy released by 1 kg of water, per 1 ºC fall in temperature , is the specific heat capacity of water, i.e. 4200 J.

(ii) The total heat energy released by water in the experiment

= m c θ = 0.02 × 4200 × (70 – 45) = 2100 J

(iii) Heat required by the calorimeter and stirrer

= m c θ = 0.16 × c × (45 – 15) = 4.8 c

By principle of Calorimetry , heat gained = heat lost , we have

4.8 c = 2100 => c = 2100/4.8 = 437.5 Jkg-1ºC-1

Numerical – 11. A hot solid of mass 60 g at 100 ºC is placed in 100 g of water at 18 ºC. The final steady temperature recorded is 20 ºC. Find the specific heat capacity of the solid.

Solution:

Heat lost by solid = m c θ = 60 × c × (100 – 20) = 4800 c J

Heat gained by water = m c θ = 100 × 4.2 × (20 – 18) = 840 J

By the principle of Calorimetry,

Heat lost = Heat gained , we have

4800 c = 840 => c = 840/4800 = 0.175 Jg-1 ºC-1 = 0.04 cal g-1 ºC--1.

Numerical – 12. 2 kg of ice melts when a jet of steam at 100 ºC is passed through a hole drilled in a block of ice. What mass of steam was used? Given: Specific heat capacity of water = 4,200 Jkg-1 ºC-1, Specific latent heat of fusion of ice = 336 × 103 Jkg-1, Specific latent heat of vaporization of steam = 2,268 × 103 Jkg-1.

Solution:

Let mass of steam used be m,

Heat gained by ice to melt to water at 0 ºC = m L = 2 × 336 × 103 J

Heat lost by steam of mass m to condensed to water at 100 ºC and then to water at 0 ºC = m L + m c θ = m × 2668 × 103 + m × 4200 × (100 – 0) J

By the principle of Calorimetry, heat lost = heat gained , we have

m × 2668 × 103 + m × 4200 × 100 = 2 × 336 × 103 => m = 0.25 kg.

Numerical – 13. 10 g of ice at 0 ºC absorbed 5,460 J of heat to melt and change into water at 50 ºC. Calculate the specific latent heat of fusion of ice. Given specific heat capacity of water is 4,200 J kg-1 ºC-1.

Solution:

Quantity of heat required to melt the ice at 0 ºC = m L = 10 L.

Quantity of heat required to raise the temperature of water from 0 ºC to 50 ºC

= m c θ = 10 × 4.2 × (50 – 0) = 2100 J.

According to question, 10 L + 2100 = 5460 => L = 336 Jg-1.

Numerical – 14.  A piece of metal at 10 º C has a mass of 50 g. When it is immersed in a current of steam at 100 ºC, 0.7 g of steam is condensed on it. Calculate the specific heat of the metal. Given Latent heat of steam = 540 cal g -1.

Solution:

Here, t1 = 10 º C, t2 = 100 ºC, mass of metal = 50 g, mass of steam = 0.7 g, latent heat of steam = 540 cal g-1, let specific heat of metal be c, θ = t2 –t1.

Heat gained by metal = m c θ = 50 × c × (100 – 10) = 4500 c J,

Heat lost by steam = m L = 0.7 × 540 = 378 J,

By the principle of Calorimetry , heat gained = heat lost , we have

4500 c = 378 => c = 378/4500 = 0.084 cal g-1 ºC-1.

Numerical – 15. A 30 g ice cube at 0 ºC is dropped into 200 g of water at 30 ºC. Calculate the final temperature of water when the entire ice cube has melted. Given: Latent heat of ice = 80 cal g-1; specific heat capacity of water = 1 cal g-1 ºC-1.

Solution:

Here, mass of ice = 30 g, mass of water = 200 g, t1 = 0 ºC, t2 = 30 ºC, t ºC = final temperature, L = latent heat of ice = 80 cal g-1, c =

pecific heat capacity of water = 1 cal g-1.

Heat lost by 200 g of water in cooling from 30 ºC to t ºC

= m c θ = 200 × 1 × (30 – t) = 200 (30 – t),

Heat gained by ice to melt to form water at 0 ºC and heating this water to t ºC

= m L + m c θ = 30 × 80 + 30 × 1 × (t – 0) = 2400 + 30 t,

By the principle of Calorimetry, heat gained = heat lost, we have

2400 + 30 t = 200 (30 – t) = 6000 – 200t

230 t = 6000 – 2400 = 3600 => t = 3600/230 = 15.65 ºC.

Numerical – 16. 1 kg of ice at 0 ºC is being continuously heated through an electric heater of 1 kW. Assuming that all the heat is transmitted to ice, calculate the time interval in second for:

(i) ice to completely melt to water at 0 ºC,

(ii) water to get heated from 0 º C to 100 ºC, (iii) water at 100 ºC to convert into steam. [Given: Specific latent heat of ice = 3,36,000 J kg-1, Specific heat capacity of water = 4,200 J kg-1 K-1, Specific latent heat of steam = 22,60,000 J kg-1 .]

Solution:

Heat supplied by 1 kW electric heater = 1000 J s-1.

(i) Heat required to melt 1 kg of ice at 0 ºC completely to water at 0 ºC = Specific latent heat of ice = 3,36000 J

Hence time required = 336000/1000 s = 336 s.

(ii) Heat required by 1 kg of water to be heated from 0 ºC to 100 ºC = m c θ = 1 × 4200 × (100 – 0) = 420000 J

Hence time required = 420000/1000 s = 420 s.

(iii) Heat required by 1 kg of water at 100 ºC to be completely into steam = Specific latent heat of steam (vaporisation) = 2260000 J

Hence time required = 2260000/1000 s = 2260 s.

Numerical – 17.  A metal of mass 250 g is heated to a temperature of 65 ºC. It is then placed in 50 g of water at 20 ºC. The final steady temperature of water becomes 25 ºC. Neglecting the heat taken by the container, calculate the specific heat capacity of the metal.

Solution:

Here, mass of metal = 250 g, mass of water 50 g, t1 = 65 ºC, t2 = 20 ºC, t = 25 ºC, c = specific heat capacity of water = ?

Heat lost by metal = m c θ = 250 × c × (65 – 25) = 10000 c

Heat gained by water = m c θ = 50 × 4.2 × (25 – 20) = 1050

[Specific heat capacity of water = 4.2 J g-1 ºC-1]

By principle of Calorimetry, heat lost = heat gained , we have

10000 c = 1050 => c = 1050/10000 = 0.105 Jg-1 ºC-1.


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