AP SSC 10th Class Physical Science Chapter 1 Solution – Heat

AP SSC 10th Class Physical Science Chapter 1 Solution – Heat

Andhra Pradesh SSC 10th Class Physical Science Chapter 1 Heat Solution for AP SSC 10th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 10 Physical Science Textbook Solution or Study Material or APSCERT Solution Class 10 Science for AP SSC 10th exam. Here you search will end! Here in this page we have provided for all question answer for Chapter 1 Heat.

For any help regarding extra notes for AP SSC 10th Class Physics/Chemistry Chapter 1 Heat you can comment us at comment box below.

AP 10th Class Physics Chapter 1 Heat Solution

 Chapter 1: Heat

1.) What would be the final temperature of mixture of 50g of water at 20°C temperature and 50g of water at 40°C temperature?

Ans:

Here given that, the 50g of water at 20°C temperature and 50g of water at 40°C temperature are mixed and we have to find the temperature of that mixture formed. We already know that, the masses of the liquids at different temperatures is same. According to that, here the temperature of mixture becomes the average of given temperature 20°C and 40°C. Hence, the final temperature of the given mixture should be 30°C.

2.) Explain why dogs pant during hot summer days using the concept of evaporation?

Ans:

Like human being in dogs no sweat glands are present to reduce their body temperature. Because of this during sunny days dogs pant by the process of evaporation to reduce their body temperature. Since, larger is the area evaporation becomes more. Ans hence, dogs tongues is having larger area due to which molecules of water from that get evaporated during panting in order to cool that surface area. In this way, dogs pant during sunny days to reduce body temperature by the process of evaporation.

3.) Why do we get dew on the surface of a cold soft drink bottle kept in open air?

Ans:

We know that, the process of condensation in which water vapour is converted into liquid water.

As the soft drink bottle is cold that means temperature of bottle and air is different. In other words temperature of air is higher as compared to temperature of bottle. We know that, air contains water molecules, when these water molecules come in contact with the cold drink bottle they loose their kinetic energy and hence become converted into liquid water. In this way the process of condensation takes place. Hence, we get dew on the surface of a cold soft drink bottle kept in open air.

 4.) Write the difference between evaporation and boiling?

Ans:

The process of evaporation and boiling both are different from each other in some aspects which are explained as given below.

Evaporation:

  • Evaporation is the process in which liquid molecules from the surface of a liquid at any temperature get evaporated.
  • Evaporation is the process in which liquid molecules get converted into gaseous molecules because of temperature or pressure difference.
  • We can observe the process of evaporation on the surface of liquid only.
  • As the process of evaporation is natural process hence it is a slow process.
  • During evaporation we cannot observe the bubbling of liquid molecules.

Example:

During sunny days, in rivers surface water get heated and because of that the water molecules from the surface get evaporated which is a natural and slow process.

Boiling:

  • Boiling is the process which requires heat which is provided externally for the process of evaporation.
  • As external heat is required for the process of boiling it is not natural process.
  • When we are boiling a liquid by giving heat to it that means only surface of that liquid is not getting heated but the whole liquid gets heat energy and starts boiling.
  • As the whole liquid involves in the process of boiling we can observe the bubbling process in boiling.
  • As boiling is not natural process and it requires external heat energy that means it is a faster process than evaporation. If heat energy given is more liquid starts boiling faster.

Example:

When we are boiling a water that time we are giving external heat energy. Because of heat given that whole water starts boiling after reaching to 100°C temperature.

5.) Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain.

Ans:

  • As we already know that, gas molecules has higher energy than liquid and solid molecules. And here the vapour phase of H2O means gaseous state of water and that gas molecules are having more energy.
  • During the process of condensation the gaseous water molecules get converted into liquid molecules. And hence gas molecules releases energy while converting into liquid molecules.
  • And the energy released is given out to the surrounding obviously due to which surrounding air become warm.
  • Thus, the surrounding air become warm when vapour phase of H2O condenses.

 6.) Answer the following questions.

a) How much energy is transferred when 1g of boiling water at 100°C condenses to water at 100°C?

Ans:

As we know that, the amount of heat energy in joules required to convert a 1 kg of liquid at its boiling point into its vapour or gaseous form at a constant temperature is called as latent heat of vaporization.

When 1gm of boiling water at 100°C condenses to water at 100°C energy is released and the energy released is called as latent heat of vaporization.

When water condenses the amount of latent heat energy released per gram is 2257J.

Hence, when 1gm of boiling water at 100°C condenses to water at 100°C the energy released is about 2257J.

b) How much energy is transferred when 1gm of boiling water at 100°C cools to water at 0°C?

Ans:

During this conversion two stages comes as given below.

When water vapours from boiling water at 100°C converts into water at 100°C releases energy which is called as latent heat of vaporization and which is 540 cal.

Now, in second stage when boiling water at 100°C converts into water at 0°C the amount of energy released is 100 cal.

Hence, the total amount of energy transferred when 1gm of boiling water at 100°C cools to water at 0°C is (540+100)= 640 cal

c) How much energy is released or absorbed when 1gm of water at 0°C freezes to ice at 0°C?

Ans:

The amount of heat energy released or absorbed when solid form is converted into its liquid form is called as called as latent heat of fusion.

The energy released during the process of conversion of water at 0°C into ice is same as the amount of energy when ice starts melting and converted into water form.

Thus, the amount of energy released or absorbed when 1gm of water at 0°C freezes to ice at 0°C is found to be 334J per gram.

d) How much energy is released or absorbed when 1gm of steam at 100°C turns to ice at 0°C?

Ans:

Here, we have to find the total amount of energy released.

Let us consider Q1 be the  energy released when water vapour at 100°C converts into water at 100°C. Hence, Q1=540 cal

Let us consider Q2 be the energy released when water at 100°C converts into water at 0°C.

Hence, Q2= 100 cal

Let us consider Q3 is the amount of energy released when water at 0°C converts into ice at 0°C . Hence Q3=  80 cal.

Now total energy released when 1gm of steam at 100°C turns to ice at 0°C is Q.

Q = Q1 + Q2 + Q3

Q = 540 + 100 + 80

Q = 720 cal

 7.) Explain the process of finding specific heat of solid experimentally?

Ans:

  • The specific heat of the substance is the amount of heat energy which is required to increase the temperature of given unit mass of the substance through 1°C.
  • And it is given by formula,
  • S = Q/m∆T
  • Now, the process of finding specific heat of solid experimentally is given as below.
  • The materials required for finding specific heat of solid are calorimeter, thermometer, stirrer, water, steam heater, wooden box and lead shots or iron bolt.
  • Now, we shall see the procedure to find specific heat of solid.
  • Initially, we will measure the mass of calorimeter along with stirrer and let us suppose it is M1.
  • Now, we fill one third part of calorimeter with the help of water and after that we will measure corresponding mass and temperature.
  • Let us consider the mass of calorimeter with water now be M2.
  • Hence, mass of water can be calculated as (M2 – M1).
  • Let us take T1 be the temperature of the water filled in calorimeter.
  • Now, we take some lead shots and insert into hot water or steam heater and we heat them upto a temperature of about 100°C. And let us consider that temperature be T2.
  • After heating this lead shots we insert suddenly them into calorimeter without loosing energy.
  • After inserting lead shots we can see the mixture get settled to a particular temperature. At this moment, let us consider the temperature be T3 and the mass of the calorimeter is M3.
  • Hence, the mass of lead shots is found to be (M3-M2).
  • As we inserted lead shots within calorimeter without loosing energy hence we can say the total heat energy is given to the calorimeter which is used by water to reach a particular temperature.
  • Now let us consider the specific heat of calorimeter, lead shots and water be Sc, Sl and Sw respectively.
  • Now by using following formula we can find the heat energy lost by the solid
  • Heat lost by the solid = (Heat absorbed by the calorimeter + Heat absorbed by the water)
  • Thus,
  • (M3-M2)×Sl×(T2-T3) = M1×Sc×(T3-T1) + (M2-M1)×Sw×(T3-T1)
  • Thus, Sl = [M1×Sc+(M2-M1)×Sw]×(T3-T1)/(M3-M2)×(T2-T3)

 

  • Thus, by using above formula and by knowing specific heat of water and calorimeter we can find specific heat of solid which is lead shots taken here.
  • This is the whole procedure to find the specific heat of solid experimentally.

8.) Convert 20°C into Kelvin scale?

Ans:

We know that the standard formula for the conversion of temperature from °C to Kelvin as given below.

T(in K) = T (in°C) + 273.15

= 20 + 273.15

= 293.15K

Thus, 20°C in Kelvin scale is found to be 293.15

 9.) Your friend is asked to differentiate between evaporation and boiling. What questions could you ask to make him to know the difference between evaporation and boiling?

Ans:

I will ask the following questions.

1) how you can define evaporation and boiling.

2) Evaporation and boiling are natural process or not?

3) Give real life example of evaporation and boiling.

4) What is the main difference you observed between the examples of evaporation and boiling in real life?

5) Our cloths get dried by which process?

6) During sunny days lakes become dried because of what?

 10.) What happens to the water when wet cloths dry?

Ans:

Generally we dry our cloths by putting or spreading them on a wire so that they will get some sunlight. Now, because of sunlight some amount of heat energy is absorbed by the water molecules present in wet cloths. Because of this heating water molecules get converted into water vapour and given to the surrounding. These water vapours cannot be seen through air. In this way our cloths get dried by the process of evaporation.

11.) Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why?

Ans:

We know that, the process of conversion of water into water vapour is called as evaporation.

And also the important point is the rate of  process of  evaporation increases as the surface area is larger or more.

If we are taking same amount of water in cup and dish we can see that the surface area is more in case of dish. And hence the water from the dish get more evaporated. And hence water from dish evaporate faster.

12.) Suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area and vapour already present in surrounding air?

Ans:

As we know that, the rate of evaporation of liquid depends on its surface area of the material and the vapour present in the surrounding air also. Now we can prove this experimentally as given below.

To prove rate of evaporation of liquid depends on its surface area:

  • First take a cup and a dish. Fill both cup and dish with same amount of water.
  • Now put them in sunlight and observe the quantity of water present in the cup and dish after 1-2 hours.
  • We can see that, the quantity of water present in dish is less as compared to the quantity of water present in cup.
  • Although we kept both in sunlight with same amount of water then also the evaporation is found to be more in dish than cup.
  • So, it happened only because of the larger surface of dish due which the water surface also gets more area and get evaporated easily.
  • But in cup so small open surface area is there due which more water surface does not come in contact with the surrounding and hence the evaporation becomes less in case of water present in cup.
  • Thus, from this activity we can conclude that the rate of evaporation of liquid is directly proportional to its surface area present. That means larger the surface area, larger will be the evaporation.

 

To prove rate of evaporation of liquid depends on the vapour present in surrounding:

  • As we did the above activity in the same we have to do the following activity.
  • Now, take a dish and fill it with some amount of water and keep it in sunlight (less humid day).
  • And observe the level of water after particular period of time.
  • Now, take the same dish and fill it with the same amount of water as you took last time and keep it in surrounding when humidity in air is more than the day before you kept.
  • And observe the level of water for the same period of time as did last time.
  • You can see that, the evaporation of water is more at less humid day than at more humid day.
  • This happened only because of the vapour present in the surrounding. On less humid day the vapour present in air are less and on more humid day the vapour present in air are also more.
  • So, from this we conclude that, the amount vapour present in air decreases the rate of evaporation of liquid.
  • Thus, the rate of evaporation of liquid is inversely proportional to the amount of vapour present in the air.

13.) Place a Pyrex glass funnel with its mouth down in a sauce pan full of water, in such a way that the steam tube of the funnel is above the water or pointing upward into the air. Rest the edge of the bottom portion of the funnel on a nail or on a coin so that water can get under it. Place the pan on stove and heat it till it begins to boil. Where do the bubbles form first? Why? Can you explain how a natural geyser works using this experience.

Ans:

  • As explained in the question if we made an arrangement as given the question we will observe the following thing.
  • We can see that, when we are giving heat to the pan after some time the bubbles are coming out from the tube like part of the funnel which is in upper side towards the air.
  • Now we can observe that, we are giving heat energy to the whole pan filled with water but initially steam is created or bubbles are coming out from upper part of the funnel. This is only because of the pressure created by heat energy given to the pan.
  • Now, when we are giving heat energy to the pan inside the funnel the water get heated and the pressure created is more than outside the funnel in pan. Because of this, at the tube like structure of funnel the bubbles are coming faster as there pressure created is more. And in this way, because of pressure created due to heating the bubbles are coming out like a geyser.

 

  • From the above experiment we can see how natural geyser works as follows:
  • Natural geysers are nothing but the source of hot water under the earth’s crust. There are many layers are present inside the earth crust. When the temperature of the layer at the bottom increases the heat energy is created which is given to the upper layer of the water present inside the earth surface. And in the same way the more pressure is created due to which temperature increases and water starts boiling.
  • And the heated water comes out from small holes with the high pressure. All this is about the natural geyser and how they works.

14.) How do you appreciate the role of higher specific heat of water in stabilising atmospheric temperature during winter and summer season?

Ans:

  • As we know that, water has high specific heat capacity that means it requires more heat to get heated. In other words it remains cool for long period of time.
  • And because of this property of high specific heat capacity of water it helps in maintaining atmospheric temperature in winter and summer days also.
  • In winter season, during the day time the water from the oceans, rivers and ponds absorbs the heat from the surrounding and during night time gives out the heat energy absorbed in order to keep the atmosphere warm during winter nights. And keeps the people warm by avoiding cold.
  • In the same manner, during summer days because of high temperature of the surrounding the water surface of rivers, ponds, oceans, swimming tank absorbs more heat energy from the surrounding and get heated.
  • Because of this more evaporation takes place. And the vapour with high energy get emitted which again leaves the atmosphere there by keeping water sources and the atmosphere cool.
  • From this we can say that, as water take more time to get heated and hence the water of the swimming tank, rivers, ponds feel like cold than the atmosphere. And we want stay in water for long time during sunny days.
  • In this way, high specific heat of water helps in stabilising atmospheric temperature during winter and summer seasons.

15.)

What role does specific heat plays in keeping a watermelon cool for a long time after removing it from a fridge at hot day?

Ans:

As we already know that, the specific heat of water is more due to which .ore heat energy or temperature is required to raise the temperature of the water.

Inside the watermelon the amount of water present is also more. When we are removing watermelon from fridge and keeping it outside for long time then also it remains cool because when we are keeping it inside the fridge the watermelon gets cooled more and after removing it from the fridge it becomes cool for some time. Also, because of high specific heat of water present inside the watermelon it takes more time to absorb heat energy from the surrounding and hence it remains cool for long time.

 16.) If you are chilly outside the shower stall, why do you feel warm after the bath if you stay in the bathroom?

Ans:

  • As we know that, the atmosphere inside the bathroom is somewhat humid than the surrounding outside. That means the number of water molecules present in the air inside the bathroom are more as compared to the water molecules present in the surrounding.
  • After taking bath when we are drying ourselves with towel that time the water molecules present inside the air of bathroom get condensed on our skin by giving heat energy outside.
  • And because of this the air or atmosphere or surrounding inside the bathroom becomes warmer and we feel warm after bath if we are staying in bathroom.
  • Condensation of water molecules is the only reason because of which we feel warm outside the chilly shower stall.

17.) Three objects, A at 30°C, B at 303K and C at 420K are in thermal contact. Then answer the following questions.

1) Which are in thermal equilibrium among A, B and C.

2) From which object to another heat is transferred?

Ans:

1)

From given data we can write as follows.

303K – 273K + 30K = 0°C + 30°C = 30°C

Thus, we can say here that objects A and B are in thermal equilibrium.

2) As we discussed above, we can say here that heat is transferred from object  C to objects A and B

Fill in the blanks.

Ans:

1.) The SI unit of specific heat is J/Kg-K.

2.) Heat flows from a body at high temperature to a body at lower temperature.

3.) Evaporation is a cooling process.

4.) An object A at 10°C and another object B at at 10K are kept in contact, then heat will flow from A to B.

5.) The latent heat of fusion of ice is 80 cal/gm.

6.) Temperature of a body is directly proportional to average kinetic energy of the molecules of the body.

7.) According to the principal of method of mixtures, the net heat lost by the hot bodies is equal to net heat gained by the cold bodies.

8.) The sultriness in summer days is due to high humidity.

9.) Water is used as a coolant.

10.) Ice floats on water because the density of ice is less than that of water.

Multiple Choice Questions:

1.) Which of the following is a warming process.

a) evaporation

b) condensation

c) boiling

d) all the above

Ans: b) condensation

2.) Melting is the process in which solid phase changes to

a) liquid phase

b) liquid phase at constant temperature

c) gaseous phase

d) any phase

Ans: b) liquid phase at constant temperature.

3.) Three bodies, A, B and C are in thermal equilibrium. The temperature of B is 45°C. Then the temperature of the C is

a) 45°C

b) 50°C

c) 40°C

d) any temperature

Ans: a) 45°C

4.) The temperature of steel rod is 330K. It’s temperature in °C is

a) 55°C

b) 57°C

c) 59°C

d) 53°C

Ans: b) 57°C

5.) Specific heat S=

a) Q/∆t

b) Q∆t

c) Q/m∆t

d) m∆t/Q

Ans: Q/m∆t

6.) Boiling point of water at normal atmospheric pressure is

a) 0°C

b) 100°C

c) 110°C

d) -5°C

Ans: b) 100°C

7.) When ice melts, it’s temperature

a) remains constant

b) increases

c) decreases

d) cannot say

Ans: remains constant

Solution by another Teacher

(1) What would be the final temperature of a mixture of 50g of water at 20 temperature and 50g of water at 40 temperature?(ASI)

Ans:-We know that the final temperature of any two mixtures,

T= (m1×t1 + m2×t2)/(m1+m2)

Or, T= (50×20 + 50×40)/(50 + 50)

Or, T=30℃

So the temperature of these two mixture is 30℃.

 

(2) Explain why dogs pant during the hot summer day using the concept 0f evaporation?(ASI)

Ans:- Dogs pant during hot summer days to reduce their body temperature. As we know dog tongue surface area is large and for this reason, the water molecules present on it evaporate.

 

(3) Why do we get dew on the surface of a cold soft drink bottle kept in the open air? (AS1)

Ans:- Because of condensation, we noticed the dew on the surface of the cold drinks bottle. As the condensation process converts the vapor into a liquid the dew is seen on cold drinks surface.

 

(4) Write the differences between evaporation and boiling? (AS1)

Ans:

Evaporation

Boiling
As compared to boiling it is a slow process

Whereas boiling is fast or rapid process

It occurs at any temperature.

But boiling needs A specific temperature.

 

(5) Does the surrounding air become warm or cool when the vapor phase of H2O condenses? Explain.

Ans:-During condensation the temperature of surrounding air became warm as the vapor molecules losses its energy to the surrounding. As a result of the temperature of vapor decreases, the temperature of the surroundings increases.

 

(6) Answer these. (AS1)

(a) How much energy is transferred when 1gm of boiling water at 100 condenses to water at100?

Ans:-Although the temperature changes are constant energy that is changing their state.This energy is called latent heat. We know that the latent heat of vaporization is 2257j/g. So the required energy to transfer 1mg of boiling water to condense is 2257 j.

(b) How much energy is transferred when 1gm of boiling water at 100 cools to water at 0?

Ans: At first we have to count the heat of vaporization as 1mg of water transfer to 100℃or gaseous state. As we know the heat of vaporization of water is 540cal/gm. And in the next, we have to calculate the heat of condensation as it convertsfrom 100℃ to 0℃.From the formula of calculation of heat Q = ms (T2-T1) (where m=mass; s is specific heat)

Or, Q=1×1× (100-0)[s=specific heat of water =1]

=100 Cal/gm

So the total energy required is (540+100) =640cal/gm.

(c) How much energy is released or absorbed when 1gm of water at 0 freezes to ice at 0?

Ans: As we know the latent heat of fusion of water is 334 j/gm.So the energy release when 1mg of water to freezes to ice is 334 j/m.

(d) How much energy is released or absorbed when 1 gm of steam at 100 turns to ice at 0?

Ans:-At first the water would vaporized. We all know that the latent heat of vaporization of water is 540. Then the energy required to transfer 100℃ to cool it at 0℃ is 100cal (Q = ms∆t).

As we all know latent heat of change of water into ice is 80cal/ gm.

So the total energy is (540 + 100 + 80)= 720 Cal/gm.

 

(7) Explain the procedure of finding specific heat of solid experimentally. (AS1)

Ans: –To calculate the specific heat of solid the following steps are required,

  • At first we have to measure the mass of the calorimeter.
  • 1/3 volume of the calorimeter must be filled with water.
  • After taking somelead (solid) shot on the calorimeter we will see mixtures came to a certain temperature.

If we calculate the mass of the temperature and mass of the calorimeter then we will see that no loss of heat to surrounding. So by this conclusion we can say heat loss by body = heat gain by water + heat gain by calorimeter.If we know the specific heat of water and calorimeter we can calculate the body also.

 

(8) Covert 20 into Kelvin scale.(AS1)

Ans:- In kelvin scale temperature will be T= 273 + 20 = 293k

 

(9) Your friend is asked to differentiate between evaporation and boiling. What questions could you ask to make him to know the differences between and boiling? (AS2)

Ans: I have to ask my friend’s about the definition of evaporation and boiling. Evaporation is when the water turns into gases at any temperature. And in Boiling, we have to give some specific temperatures.

 

(10) What happens to the water when wet clothes dry? (AS3)

Ans: Water will be evaporated when wet clothes dry.

 

(11) Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why?(AS3)

Ans: As the dish has more wide area than the cap the water will be evaporated more in the dish. As we all have known that in more surface area more evaporation takes place.

(12) Suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area and vapour already present in the surrounding air. (AS3)

Ans: If we take two dishes one has a larger surface area than the others we will find the dishes which has a larger surface area evaporated first. As the evaporation increase with the increase of area.

 

(13) Place a Pyrex funnel with its mouth-down in a saucepan full of water, in such a way that the stem tube of the funnel is above the water or pointing upward into air. Rest the edge of the bottom portion of the funnel on a nail or on a coin so that water can get under it. Place the pan on a stove and heat it till it begins to boil. Where do the bubbles form first? Why? Can you explain how a geyser works using this experience? (AS4)

Ans:-The bubbles will form at the bottom of the saucepan where the nail or coin is placed. As the temperature increases, the pressure will be increased so this phenomenon will happen. At the bottom of the geyser, the boiling will be started. As a result, the eruption of the bubble will begin on the above. So this geyser works like a natural geyser.

(14) Collect information about working of geyser and prepare a report. (AS4)

Ans: –The working of geysers is actually the conversion of electrical energy into heat energy. Some element is present in the geyser which heated up. There is a separate inlet and outlet of cold and hot water in the geyser.

(15) Assume that heat is being supplied continuously to 2kg of ice at -5. You know that ice melts at 0 and boils at 100 Continue the heating till it starts boiling. Note the temperature every minute. Draw a graph between temperature and time using the values you get. What do you understand from the graph? Write the conclusions. (AS5)

Ans:

From the graph, we concluded that if the melting of ice will complete then the temperature will be changed. If the latent heat is absorbed and the fusion is absorbed then no changes of temperature are observed.

 

(16) How do you appreciate the role of the higher specific heat of water in stabilizing atmospheric temperature during winter and summer seasons? (AS6)

Ans: –As the specific heat of water is high there is seen stability in the increase in temperature difference in the summer and winter season. Although the equator gets more heat from the sun the ocean water in that region doesn’t absorb more heat due to high specific heat. These conditions also help to maintain the temperature of the air during summer as well as winter.

 

(17) Suppose that 1l of water is heated for a certain time to rise and its temperature by 2. If 2l of water is heated for the same time, by how much will its temperature rise? (AS7)

Ans:-All we know that Q = ms∆t

So in these two case Q must be the same. ∆t2 = (m1×∆t1/m2) = (1×2/2) =1

So if 2l of water is heated then 1℃ temperature will be increased.

 

(18) 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:-The specific heat of water is very high. As in watermelon more water inside it cools for a long time.

 

(19) If you are chilly outside the shower stall, why do you feel warm after the bath if you stay in the bathroom? (AS7)

Ans: –After staying sometime when we came out from the bathroom we feel warm as there is more vapor outside. As the vapor molecules present more as compared to the bathroom in outside warm-weather feel.

FILL IN THE BLANKS

(1) The SI unit of specific heat is______joule_______

 

(2)  __________Heat___ flows from a body at higher temperature to a body at lower temperature.

 

(3) ____Freezing____ is a cooling process.

 

(4) An object ‘A’ at 10 and another object ‘B’ at 10K are kept in contact, then heat will flow from ___A______ to ________B___.

 

(5) The latent heat of fusion of ice is ____33600 j/k________.

 

(6) Temperature of a body is directly proportional to ________ thermal energy_____.

 

(7) According to the principle of method of mixtures, the net heat lostby the hot bodies is equal to______heat gained_____ by the cold bodies.

 

(8) The sultriness in summer days is due to

____vaporization _________.

 

(9) ____water________ is used as a coolant.

 

(10) Ice floats on water because ___ low-density _________.

 

MULTIPLE-CHOICE QUESTIONS

(1) Which of the following is a warming process –

(c) Boiling.

 

(2) Melting is a process in which solid phase changes to –

(a) Liquid state.

 

(3) Three bodies A, B and C are in thermal equilibrium. The temperature of B is 45 then the temperature of C is- is

(a) 45℃.

 

(4) The temperature of a steel rod is 330K. Its temperature in

(b) 57℃

 

(5) Specific heat S = –

(c) Q/m∆t.

 

(6) Boiling point of water at normal atmospheric pressure –

(b) 100℃.

 

(7) When ice melts, its temperature –

(a) Remain constant.

Updated: August 26, 2023 — 2:37 pm

Leave a Reply

Your email address will not be published. Required fields are marked *