Telangana SCERT Class 9 Physical Science Chapter 11 Solution – Heat. Here in this post we provides Class 9 Physical Science Heat Telangana State Board Solution. Telangana State Board English Class IX Medium Students can download this Solution to Solve out Improve Your Learning Questions and Answers.
Telangana State Board Class 9 Physical Science Chapter 11 Heat Solution:
CHAPTER: – 11
Reflections on concepts
1) Why do we get dew on the surface of a cold soft drink bottle kept in open air?
Ans: – In condensation, there is seen the dew on the surface of the cold drinks bottle. For condensation process converts this vapour into a liquid the dew is seen on cold drinks surface.
2)Water can evaporate at any temperature Explain with an example?
Ans: – The water has a specific boiling and vaporisation temperature after the surplus of that temperature we see that water is converted to vapour which is 100℃ as after this temperature water vaporisation started.
3)What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day? (AS7)
Ans: – As we all know that the specific energy is more that the other liquid so because of this advantage the water inside the watermelon kept it cool all the time. Its nothing the benefits of specific heat of water.
4)Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why? (AS3)
Ans: – Dish has more wide area as compared to the cap so,the water will be evaporated more in the dish. It is known to us that in more surface area more evaporation takes place.
5)Why specific heat is different for different substances? Explain (AS1)
Ans: – The specific heat is different in different substance as the boiling and vaporization temperature of different object are different. As from the equation we know that Q=MS∆T, so it clearly sees that the specific heat depends on different things like masses, temperature difference and heat capacity.
Application of concepts
1) Using the concept of evaporation explain why dogs pant during hot summer days? (AS1)
Ans: – As we all the evaporation of water from the surface of the body makes the body cools so as the f9gs pant during hot summer days ut reduces its body temperature by doing this.
2)If 50g of water at 20°C temperature and 50 g of water 40°C temperature is mixed, what is the final temperature of the mixture of?
Ans: – The final temperature of any two mixtures is determined by,
T= (m1×t1 + m2×t2)/(m1+m2).
Here, m1=m2=50kg, T1=20℃ and T2=40℃.
So, T= (50×20 + 50×40)/(50 + 50)
Then the temperature of these two mixtures is 30℃.
3)What do you observe in the surroundings in terms of cooling or heating when water vapour is getting condensed (AS1)
Ans: – The condensation always kept the temperature control as in this process the water vapour after coming out from the body or the surrounding make the area or the places cool in nature. So, the vaporization is essential for the reducing the temperature in the surrounding.
Higher Order Thinking Questions
Answer these (AS1)
(a) How much energy is transferred when 1 g. of boiling water at 100°C condenses to water at 100°C?
Ans: – This energy is called latent heat.
As the latent heat of vaporization is 2257j/g.
The required energy to transfer 1mg of boiling water to condense will be 2257 j.
(b) How much energy is transferred when 1 g. of boiling water 100°C cools to water at 0°C?
Ans: – The heat of vaporization of 1mg of water transfer to 100℃or gaseous state and heat of vaporization of water is 540cal/gm.
For calculate the heat of condensation as it converts from 100℃ to 0℃.
As the heat, Q = ms (T2-T1) (where m=mass; s is specific heat)
Or, Q=1×1× (100-0)[s=specific heat of water =1]
Required energy is (540+100) =640cal/gm.
(c) How much energy is released or absorbed when 1 g. of water at 0°C freezes to ice at0°C?
Ans: – The latent heat of fusion of water is 334 j/gm.
Then energy release when 1mg of water to freezes to ice is 334 j/m.
(d)How much energy released or absorbed when 1 g. of steam at 100°C cools to ice at 0°C?
Ans: – The latent heat of vaporization of water is 540 and energy required to transfer 100℃ to cool it at 0℃ is 100cal (Q = ms∆t).
The latent heat of change of water into ice is 80cal/ gm.
Then the total energy is (540 + 100 + 80)= 720 Cal/gm.
2)Suppose that 1 L of water is heated for a certain time to rise its temperature for 2°C. If 2L of water is heated for the same time, how much of its temperature would rise?(AS7)
Ans: – As, Q=ms∆t
Q must be the same in these cases. Q1=Q2.
So, ∆t2= (m1×∆t1/m2) = (1×2/2) =1
When 2l of water is heated up 1℃temperatures will be increased in that case.
MULTIPLE CHOICE QUESTIONS.
1.) Which of the following is a warming process
a) Evaporation b) condensation c) boiling d) all the above.
Ans: – Option (c).
2.) Melting is a process in which solid phase changes to
a) liquid phase b) liquid phase at constant temperature c) gaseous phase d) Gaseous phase at constant temperature.
Ans: – option (a).
3.) Three bodies A, B and C are in thermal equilibrium. The temperature of B is 45°C. then the temperature of C is
a) 45°C b) 50°C c) 40°C d) any temperature.
Ans: – Option (a).
4.) The temperature of a steel rod is 330K. Its temperature in °C is
a) 55°C b) 57°C c) 59°C d) 53°C.
Ans: – option (b).
5.) When ice melts, its temperature
a) remains constant b) increases
c) decreases d) first decrease and then increase.
Ans: – option (a).