# Maharashtra Board Class 10 Science Part – 1 Chapter 5 Heat Solution

## Maharashtra Board Class 10 Science Part – 1 Solution Chapter 5 – Heat

Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 5: Heat. Marathi or English Medium Students of Class 10 get here Heat full Exercise Solution.

 Std Maharashtra Class 10 Subject Science Part – 1 Solution Chapter Heat

### Chapter: -5 (Heat)

Exercise: –

1) Fill in the blanks and rewrite the sentence.

a) The amount of water vapour in air is determined in terms of its ……absolute density……

b) If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their ……different specific heat capacity…………

c) During transformation of liquid phase to solid phase, the latent heat is ………latent heat of fusion….

2) Observe the following graph. Considering the change in volume of water as its temperature is raised from 0℃, discuss the difference in the behavior of water and other substances. What is this behavior of water called?

Ans: – From the graph we see that water reaches its maximum density at 4℃ and after that it shows normal behaviour like others. This is a specialty of water. Normally any element cooling when it temperature decrease but in water it sometimes different.

3) What is meant by specific heat capacity? How will you prove experimentally that different substances have different specific heat capacities?

Ans: – Specific heat is the amount of heat required to raise the temperature of an object of unite mass by 1℃.

If we rake three different solid ball of lead, iron and copper and after heatig at 100℃ when we placed them in wax, then we will see that iron ball comes more deepest comparing other two. So we can tell that iron has more specific heat capacity as heat absorb depends on specific heat.

4) While deciding the unit for heat, which temperatures interval is chosen? Why?

Ans: – The time intervals are 14.5℃ – 15.5℃ while deciding the unite for heat.

This value is chosen as when we measure the specific heat of water which is 1 cal/g.℃, then we see that the temperature is varies between 14.5 – 15.5℃.

5.) Explain the following temperature versus time graph.

Ans: –  From the graph we clearly see that the temperature is changing in various state of water. Line AB represent that the temperature is constant and ice is converting to water. After heating up the line shows is BC which represent that temperature is rising up to 100℃ and water after boiling transfer to vapour. As we all know that after 100℃ water become vapour.

6.) Explain the following:

a.) What is the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?

Ans: – Water shows different property as it density is at peak at 4℃. At from 0℃ to 4℃ water density rises and after that it density decrease this property of water is anomalous behaviour helps to preserving aquatic life in regions of cold climate.

b.) How can you relate the formation of water droplets on the outer surface of a bottle taken out of refrigerator with formation of dew?

Ans: –Water vapour condense when it kept from freezing temperature to higher temperature for sometimes. In this case the bottle after some time when we taken out from the refrigerator we see the dew on the surface because of the temperature difference as water vapour condense outside.

c.) In cold regions in winter, the rocks crack due to anomalous expansion of water.

Ans: –Water shows anomalous behavior at 0 ℃ to 4℃ temperature. From zero to 4-degree water density increase at 4℃ it maximum. From this temperature difference in cold region rock volume increasing continuously.

a.) What is meant by latent heat? How will the state of matter transform if latent heat is given off?

Ans: – Latent heat is the constant temperature at which any substance changes its states from one form to other.

The bond strength between the atoms and molecules are affected by latent heat as while changing from one state to another latent heat is constant. If latent heat given off, then bond strength changes.

b.) Which principle is used to measure the specific heat capacity of a substance?

Ans: –  The calorimetry method is the main principal to measure the specific heat capacity of a substance.

c.) Explain the role of latent heat in the change of state of a substance?

Ans: – In every substance potential energy is stored in them. The particles of the substance are bonded with the help of this potential energy as when potential energy changes the substance change its state. The latent heat supplied this potential energy to the substance.

d.) On what basis and how will you determine whether air is saturated with vapour or not?

Ans: – Water vapour percentage in air determine the whether air is saturated or unsaturated.  When the vapour amount is more than the air can contain it said saturated and when it is lower than the air contain is unsaturated situation.

If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy. The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.

i.) Heat is transferred from where to where?

Ans: – After reading the upper paragraph we can say that heat is transferred from the higher temperature to the lower temperature of an object.

ii) Which principle do we learn about from this process?

Ans: – The principal of heat exchange process is the main principal of which we learn from it.

iii.) How will you state the principle briefly?

Ans: – We can say that in the principal of heat exchange process the system is isolated. The heat gain by lower temperature object is equal to heat release by higher temperature object.

iv.)  Which property of the substance is measured using this principle?

Ans: – We can measure the specific heat of an object by using this principal.

9.) Solve the following problems:

a.) Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3 0C and B by 5 0C. Which object has more specific heat? And by what factor?

Ans: – we know that specific heat of an object s =∆Q/m×∆t. where Qis heat of the object, m is the mass and ∆t is temperature difference.

For object A, S1= Q/1×3 = Q/3 cal/g℃. (As both object heat is Q which is same).

For object B, S2= Q/ 1× 5 = Q/5 cal/g.℃.

So, s1/s2 = Q/3/Q/5 = 5/3.

So object A has more specific heat by factor 5/3.

b.) Liquid ammonia is used in ice factory for making ice from water. If water at 20 0C is to be converted into 2 kg ice at 0 0C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia= 341 cal/g)

Ans: – Heat release by object of mass 2kg from 20℃ to 0℃ is (2×1×1000×20 = 4000) 40000 cal.

Heat release by object to become ice from 0℃ is (2×1×1000×80=160000 cal.

Total heat release is (160000 + 40000 = 200000) 200000 cal.

So the amount of ammonia which will be evaporated is (200000/341 = 586.5)586.5 g.

c.) A thermally insulated pot has 150 g ice at temperature 0 0C. How much steam of 100 0C has to be mixed to it, so that water of temperature 50 0C will be obtained?

(Given: latent heat of melting of ice = 80 cal/g, latent heat of

vaporization of water = 540 cal/g, specific heat of water = 1 cal/g 0C)

Ans: –  Heat required for the object of mass 150g for converting to ice is = 150 × 80 = 12000 cal.

Heat required for heating up to 50℃ is = 150 ×1×50= 7500 cal.

Total heat required is 12000 + 7500 = 19500 cal.

Let m amount of stem will be mixed.

Total heat release to convert m g of steam is = heat release from 100 ℃ to water + heat release from 100℃ to water at50℃ = (540m + 50m) =590m.

From the principal of calorimetry, 19500 = 590 m; or, m=19500/590 = 33g.

d.) A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg 0C. It contains 250 gm of liquid at 30 0C having specific heat of 0.4 kcal/kg 0C. If we drop a piece of ice of mass 10 g at 0 0C, what will be the temperature of the mixture?

Ans: – let the temperature of the mixture be T.

Heat required to convert to ice of 10g mass is =10×80= 800 cal.

Heat required to become at T℃ of water is = 10×1×T = 10T.

Total heat required is 800 cal + 10T.

Calorimeter heat release at 30℃ to ice is = 100×0.1×(30-T) =10(30-T).

Heat release by 250g of liquid at 30℃ is = 250 ×0.4×(30-T) =100(30-T).

Heat release in the total processing is {100(30-T) + 10(30-T)} =110(30-T).

Applying calorimetry principal,800 + 10T = 110(30-T)

Or, T = 20.83℃.

Then the temperature of the mixture will be 20.83℃.

Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 5 Heat

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Updated: December 1, 2021 — 1:05 pm