Physics

Effect of Heat on Matter

Chapter- 6

Q. 1. At the time of construction of a rail line why are small gaps kept in between two rails?

Ans: To avoid bending of the rails due to its thermal expansion

Q.2. Why do we feel comfort when wind is blown by fans over our sweating body?

Ans: Evaporation prouces cooling

Q.3. With the help of latent heat - 

Ans:  Temperature of a body increases, state of a substance changes, 

         internal energy of a body increases


Heat is a type of energy.

The combined kinetic energy of molecules of a substance is called heat energy.

The SI unit of heat is Joule (J).

1 Calorie (cal) = 4.2 Joule (J).

The amount of heat energy required to increase the temperature of 1 gm of water by 1⁰C is called 1 Calorie. 

The international unit of temperature is Kelvin (K).

The property of a substance which changes with temperature and, measuring the change accurately, is called - Thermometric Property.

The thermometric properties of substance are - Volume, Resistance & Pressure.

At one atmospheric pressure and at the temperature at which ice melts is called - 0૦C.



Expansion of Solids:
Volume of almost all objects increases to some extent if they are heated. It is not difficult to understand the cause of heat and temperature if we explain them with the help of an atomic model. We can imagine a solid as a collection of many molecules. 
When the solid is heated, its molecules will vibrate. The higher the temperature, the more the molecules will vibrate.
Length, breadth and height of a solid are expand equally in three directions by the application of heat.

Q.1. When a pot of mercury is heated, the height of mercury decreases first then increases. Why?

Ans: When a pot of mercury is heated, the pot gets heated at first. As a result, it expands. The volume of the pot increases and the height of mercury decreases. But the expansion of mercury is much higher than that of glass. As a result, the mercury gets heated and the height of mercury increases gradually.

Q.2. Does temperature exist in space where there are no atoms or molecules?
Ans: Yes, temperature exists in space. Various types of radiation are nearby galaxies. And the temperature is related to any kind of radiation. Therefore, temperature obviously exists in space.

Q.3. Why do we feel cold when we come from a hot, humid, and crowded place to an open area?
Ans: The amount of water vapour is more in a humid crowded place. As a result, the sweat of body does not get dried easily. On the other hand, in an open area sweat gets dried easily because the amount of water vapour is less. The latent heat required to dry the sweat is absorbed from the body . As a result, the of body decreases and we feel cold.

The most common example of the expansion of liquid is a thermometer. There are many types of thermometers, among them perhaps the clinical thermometer is the most familiar.
A glass tube contains mercury at the bottom. If heat is applied, the volume of mercury is increased and it goes up through the capillary tube. 









Specific Heat:
The amount of heat required to increase the temperature of a body of mass 1 kg by 1 K is called the specific heat. 

That is, if an amount of heat Q is required to increase the temperature from T1 to T2, the specific heat will be:

                                S = Q/ m (T2 - T1)

                     The unit of specific heat - J kg ¯ ¹ K¯¹

Heat capacity C is the amount of heat required to increase the temperature 1 K of a body. 
The amount of heat required to increase the temperature 1 K of a body of mass 1 kg is called specific heat. So we can easily find out the heat capacity C of a body if we know the specific heat of that substance. If the mass of an body is m and the specific heat is s, then the heat capacity will be:

                                      C = ms
                      The heat capacity of 10 kg gold is
                             c = 10 x 230 JK ¯¹
                    i.e. about 20 times more.

This means that gold or any other metal can be heated quickly but water cannot be heated so easily.






Q.1. When a pot of mercury is heated, the height of mercury decreases first then increases, why?
Ans: When a pot of mercury is heated, the pot gets heated at first. As a result, it expands. The volume of the pot increases and the height of mercury decreases. But the expansion of mercury is much higher than that of glass. As a result, the mercury gets heated and the height of mercury increases gradually.

Q.2. Does temperature exist in space where there are no atoms or molecules?
Ans: Yes, temperature exists in space. Various types of radiations are present in space. The sources of these radiations are nearby galaxies. And temperature is related to any kind of radiation. Therefore, temperature obviously exists in space.

Q.3. Why do we feel cold when we come from a hot, humid and crowded place to an open area?
Ans: The amount of water vapour is more in a humid crowded place. As a result, the sweat of body does not get dried easily. On the other hand, in an open area sweat gets dried easily because the amount of water vapour is less. The latent heat required to dry the sweat is absorbed from the body. As a result, the temperature of body decreases and we feel cold.

Q.4. Why do the drops of water condense on the surface of a glass when a piece of ice is kept in the glass of water?
Ans: When a piece of ice is kept in a glass of water, its temperature decreases compared to the environment. As a result, water vapour present in air gets condensed on its surface. Thus drops of water condense on the surface of the glass.

Q.5. Why is cooking done faster in a pressure cooker?
Ans: The boiling point of water depends on pressure. The boiling point of water increases with the increase of pressure and decreases with the decrease of pressure. In a pressure cooker, the boiling point of water is increased by increasing the pressure. As a result, water reaches a higher temperature which makes cooing faster.






Physics Math:
Q.1. Scientist Celsius introduced a thermometer in which the boiling point of ice was 100⁰C and the boiling point of water was 0⁰ C! At which temperature the readings in Celsius and Fahrenheit scales are equal?

Solution:
According to the thermometer introduced by scientist Celcius the relation between Celcius and Farenheit scales is -

                                                 Tc  - 100/ 0 - 100    =  TF - 212 / 32 - 212


                                 Or,           Tc - 100      = -100 / -180 (TF  - 212 )
                                                                            = 5/9 (TF - 212)

                         If Tc and TF  are equal,
                                      9Tc  - 900   =  5Tc - 1060
                                         Or,   4Tc   = 160
                                     Tc = - 40

    Therefore, at  - 40⁰ the reading in celcius and Farenheit scales are equal.


Q.2. At which temperature will the density of gold decrease to 0.001%?
Solution:
Let, the initial and final density and volume of gold are ρ0, V0 and ρ, V respectively.

                       ρ = m/V ,    ρo  = m/Vo

     Now,     V - Vo   =  m( 1/ρ  - 1/ρo)

    or,  △V   〓  - m/ ρρo ( ρo - ρo)

     or, △V   〓  - m/ ρρo   △ρ

      Or,   △V/ Vo   = △V/ ρ0      = - ( -0.001/ 100)
           ('- ' sign indicates decrease of density) 

       or,    ∝   △ T   =  0.001 / 100
     or,     △ T = 0.001/ 100 x 3x  x  ∝      (   ∝ = 14 x 10¯⁶ )
             T2 - T1    = 0.001/ 100 x 
                       T2  =  0.238⁰ C 


Q.3. The temperature of water increased by 15⁰ C after releasing a piece of hot iron of mass 1 gm of temperature 30⁰ C was put into water of 1 liter. What was the temperature of the piece of iron?

Solution:
Given,
mass of iron, m1  = 1 gm = 0.001 kg
mass of water , m2 = mass of 1 liter water = 1 kg
specific heat of iron, S1   = 0.45 x 10³  J kg ¯¹k ¯¹
Specific heat of water , S2 = 4.2 x 10 ³ J kg ¯¹k ¯¹

Here the heat lost by the iron is equal to the heat gained by water. Let the initial temperature of the iron is TC. And the final temperature of both iron and water is = (30⁰ + 15⁰) = 45⁰C

Now      m₁S △θ = m₂S △θ

                  m₁S (T - 45)  = m₂S (45 - 30)
                 
                        T - 45 = m₂S₂  x 15 / m₁S₁  ⁰C

               T = (1 x 4.2 x 10³  x  15 / 0.001 x 0.45 x 10³     + 45 ) ⁰C
                T = 140045 ⁰C Ans.

Q.5. If the temperature of a gas confined in a closed cylinder is increased from 30⁰C to 100⁰C, by what percentage will the pressure of the gas be increased?


Solution:
Here given, initial Temperature, T₁ = 30 ⁰C = 303 K
Final Temperature, T₂ = 100⁰C = 373 K
At constant volume of gas

                            P ∝ T

P₁/T₁   = P₂/T

P₂/P₁  =  T₂/T

   P₂ - P₁/P₁  = T₂ - T₁/ T₁ x 100%
                     = 373 - 303 / 303   x 100%
                     = 23.1% Ans.


                  

                                                       
                                                 


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