10th class. HEAT LESSON

chapter 1:- Heat

short notes

Heat is a form of energy which flows from the body at higher temperature to the body that has lower temperature. (from hot body to cold body.)

Formula :- Q=ms∆t
Where Q= heat (absorbed or lost)
m = mass
∆t = temperature difference

Units :-
1. joule (J) in S.I system)
2. calorie(cal) (in C.G.S system)

3. I cal = 4.186 joules.
5. Calorie: The amount of heat required to raise the temperature of 1 gram of water by 1^°C is called calorie.

1. The degree of hotness or coldness is called temperature.
2. It depends on the average kinetic energy of the molecules of a system.

3. Units:-
a. kelvin(K) (in S.I system)
b. degree celsius(in C.G.S system)

4. Normal temperature of a healthy human is 98.6° F or 37° C or 310 K.
5. The rate of rise in temperature depends on the nature of the substance.
6. Temperature measured on the Kelvin scale is called absolute temperature.
7. 0°C = 273K
8. Temperature in Kelvin = 273 + Temperature in Celsius.
9. The average kinetic energy of the molecules is directly proportional to the absolute temperature.

If two different systems, A and B, (thermal contact) are in thermal equilibrium individually with another system C, then the systems A and B are in thermal equilibrium with each other.
1. When two bodies are in thermal equilibrium, they are at the same temperature.
2. The state of a thermal equilibrium denotes a state of body where it neither received nor gives out heat energy.
Ex:- 1. If you are not feeling either hot or cold in your surroundings, then your body is said to be in thermal equilibrium with the surrounding air.
2. The furniture in the room is in thermal equilibrium with air in the room.

Activity -3

Aim:-
To prove The average kinetic energy of the molecules is directly proportional to the absolute temperature.

Required apparatus:-
Food colour, two bowls, hot water and cold water.

Procedure:-
1. Take two bowls, one with hot water and second with cold water.
2. Gently sprinkle food colour on the surface of the water in both bowls.
3. Observe the motion of the small grains of food colour.

Observations:-
We will observe the jiggling of the grains of food colour in hot water is more than that of cold water.

Conclusion:-
The average kinetic energy of the molecules is directly proportional to the absolute temperature.

Activity -4

Aim:-
To prove that the Heat is the energy that flows from a hotter body to colder body

Required apparatus:-
Container, cylindrical transparent glass jar, coconut oil, two thermometers, water and lid etc.

Procedure:-
1. Take a cylindrical transparent glass jar and fill half of it with hot water at 60°C.
2. Very gently pour coconut oil over the surface of the water.
3. Put a lid with two holes on the top of the glass jar.
4. Take two thermometers and insert them through the holes of the lid in such a way that the bulb of the one thermometer lies inside the water and the other lies inside the coconut oil.
5. Now observe the readings of the two thermometers.

Observations:-
1.The reading of the thermometer kept in water decreases, at the same time, the reading of the thermometer kept in oil increases.
2. Water loses energy while oil gains energy. This means,water molecules kinetic energy decreases while the oil molecules kinetic energy increases.

Conclusin:-
We concluded that temperature determines the direction of heat (energy) flow, whereas, heat is the energy that flows.

Activity -5

Aim:-
To prove that the rate of rise in temperature depends on the nature of the substance.
Required apparatus:-
Large jar, Test tubes-2, oil, water, thermometers-2, retort stand, one holed corks.
Procedure:-
1.Take a large jar with water and heat it up to 80℃
2.Take two identical boiling test tubes with one holed corks.
3.Fill one test tube with 50gm of water and the other with 50gm of oil, both at room temperature.
4.Insert thermometers in test tubes through corks.
5.Place them in hot water jar with the help of stands.
6. As they kept in the same hot water jar for the same intervals of time, the same amount of heat is supplied to water and oil.
7.Now, Observe and note down the readings of thermometers for every five minutes.

Observations:-
We observe that the rate of rise in temperature of the oil is higher than that of the rise in temperature of the water.

Conclusion:-
we conclude that the rate of rise in temperature depends on the nature of the substance.

1. The amount of heat (Q) absorbed by a substance is directly proportional to its mass (m).
i.e Q ∝ m -------(1)
2. For the same mass (m) of water the change in temperature is proportional to the amount of heat (Q) absorbed by it.
i.e Q ∝ ∆T --------(2)
3. From equation (1) and (2), we get Q ∝ m ∆T
=> Q = m S ∆T
4. Where ‘S’ is a constant for a given substance. This constant is called “specific heat” of the substance.
The specific heat of a substance is the amount of heat required to raise the temperature of the unit mass of the substance by one unit.

formula:-
Specific heat $$ {S=\frac{Q}{m∆T}} $$ Where,
   Q = Amount of heat absorbed by a substance.
   m = Mass of the substance.
   S = Specific heat of the substance.
   ∆T = Change in temperature.
Units :- 
1. SI unit is $${\frac{J}{kg-K.}}$$
2. CGS unit is $${\frac{cal}{gm-℃.}}$$
Conversation:-     1 cal/gm-℃ = 4.186×10³ J/kg-K
1. The specific heat is different for different substances.
2. Water has a greater specific heat value
3. Water specific heat=1 cal/gm-℃).

1. The molecules of the system have different forms of energies such as 

a) linear kinetic energy, 

b) rotational kinetic energy, 

c) vibrational energy and 

d) potential energy 

                              between molecules.
2. The total energy of the system is called internal energy of the system..

Oceans:-  Specific capacity value of water(oceans) in stabilizing atmospheric temperature during winter and summer seasons.
Watermelon:-
Specific heat capacity plays in a watermelon to keep it cool for a long time after removing it from a fridge on a hot day.
Samosa:-
A samosa appears to be cool outside but it is hot when we eat it because the curry inside the samosa contains ingredients with higher specific heats.
Nuclear reactor:-
Water has greater specific heat values. Hence water is used as a moderator in a nuclear reactor.

Let the temperatures of two masses m1 and m2 are T1 and T2(T1 is high temperature and T2 is low temperature) then,
temperature of mixture
$$ { T=\frac{(m_1T_1+m_2T_2}{(m_1+m_2)}} $$
When two or more bodies at different temperatures are brought into thermal contact, then net heat lost by the hot bodies is equal to net heat gained by the cold bodies until they attain thermal equilibrium.
Net heat lost = Net heat gain.
This is known as the principle of the method of mixtures.

The specific heat of a solid (lead) can be calculated by finding the values experimentally and substituting them in the following formula $$ {S_l=\frac{[m_1s_c + (m_2-m_1) S_w](T_3-T_1)}{(m_3-m_2)(T_2-T_3)}} $$ m₁= mass of calorimeter
m₂-m₁= mass of water
m₃-m₂ = mass of lead shots
T₁= temperature of calorimeter and water
T₂=temperature of lead shots
T₃= temperature of calorimeter along with water and lead shots
S_l = specific heat of lead shots
S_c= specific heat of calorimeter
S_w= specific heat of water

1. The process of escaping of molecules from the surface of a liquid at any temperature and pressure is called evaporation.
2. Evaporation is a surface phenomenon(occurs at the surface)
3.When evaporation taking place kinetic energy of all molecules decreases because the molecules on top of the liquid gain energy from the molecules inside to leave the surface.
4.The temperature of liquid decreases due to the decrease in kinetic energy of molecules.
so, evaporation is a cooling process

Examples:-

1. When wet clothes dry, you will notice that water in the clothes evaporates and mixes with the surrounding air.
2. When the floor of a room is washed with water, the water on the floor evaporates within minutes and the floor becomes dry.
3. Sweating mechanism of our body is also an example of the evaporation process.
4. In hot summer days dog pants to decrease it's body temperature by evaporation.
5. The concept of evaporation involved in the pigs toiling in mud.
6. Pot has pores (small holes). Water inside the pot starts evaporation through the holes by gaining the heat energy from the pot.
Due to this evaporation process water in a pot tends to be cool.

Factors effecting the evaporation:-

Process of evaporation is effected by,
     1. Surface area.
    2. Wind speed.
    3. Humidity.
    4. Temperature.

Examples

1. The water kept in a china dish evaporates faster than in a cup because of more surface area.
2. Water in wet clothes kept under the fan evaporates faster than in normal conditions.
3. Water in wet clothes evaporates faster on a less humid day than on a more humid day.
4. Wet clothes dry faster in the summer season due to high temperatures.

Evaporation and boiling:-

In both processes the liquid state of the substance changes to gaseous state.
But, boiling takes place at a particular temperature( boiling point) and pressure only where, evaporation takes place at any of these two.
So the boiling point is fixed to a liquid but not the evaporating point.

CONDENSATION

Condensation is a reverse process of evaporation.
1. The phase change from gas to liquid is called condensation.
2. It is a warming process.
When condensation takes place the total internal energy of a liquid increases.

Examples:-

Water droplets on a cool drink bottle or a glass containing cool water are formed due to condensation of water vapour from surrounding air.
2. We feel warm after finishing our bath under the shower on a hot day due to condensation.

HUMIDITY

1. The amount of water vapour present in air is called 'humidity'.
2. The presence of vapour molecules in air is said to make the atmosphere humid.

DEW

1. During winter nights, the atmospheric temperature goes down.
2. The surface of window panes, flowers, grass etc, becomes even colder.
3. The air near them becomes saturated and condensation begins.
4. The water droplets condensed in such a surface are known as dew.

FOG

1. During the low temperature, the water molecules present in vapor condense on the dust particles in air and form small droplets of water.
2. These droplets keep floating in the air and form a thick mist. This thick mist is called fog.

1. Boiling is a process in which the liquid phase changes to gaseous phase at a constant temperature at a given pressure.
2. water boils at water 100℃ or 373K (Boiling point of water).

Aim:- To Understanding the concept of boiling:

Required apparatus:-

Beaker, burner, thermometer etc.

Procedure:-

1. Take a beaker of water, keep it on the burner.
2. Note the readings of the thermometer for every 2 minutes.

Observations:-

1. You will notice that the temperature of the water rises continuously, till it reaches 100°C.
2. Beyond 100°C no further rise of temperature of water is seen.
5. At 100℃, though supply of heat continues, the temperature does not increase further.
6. We also observe a lot of bubbling at the surface of water at 100℃.

Conclusion:-

1. This process of converting the liquid into vapour (gas) continues as long as you supply heat. This appears as water that is boiling for us.
2. Boiling takes place at a fixed temperature(100℃). This is called boiling point of water.

Latent Heat of Vaporization(L _v )

1. This heat energy is used to change the state of water from liquid to vapour (gas). This is called latent heat of vaporization.
2. Latent heat of vaporization, $$ {L_v=\frac{Q}{m}} $$
3. Latent heat of vaporization is 540 cal/gm.

Units :-

1. SI unit is J/kg.
2. CGS unit is cal/gm.

1. The process of converting solid into a liquid is called “Melting”.
2. Melting point of water melts at is at 0°C or 273K.

Aim:- To Understanding the concept of melting:

Required apparatus:-

Beaker, ice cubes, thermometer, burner etc.

Procedure:-

1. Take small ice cubes in a beaker. Insert the thermometer into ice cubes in the beaker.
2. Observe the reading of the thermometer.
3. Now start heating the beaker keeping it on a burner.
4. Observe changes in the thermometer reading every 1 minute till the ice completely melts and gets converted into water.

Observations:-

1. You will observe that the temperature of ice at the beginning is equal to or below 0℃.
2.If the temperature of ice is below 0℃,it goes on changing till it reaches 0℃.
3. When ice starts melting, you will notice no change in temperature though you are supplying heat continuously.

conclusion :-

1. During melting the temperature remains the same.
2. This process takes place at a constant temperature 0°C or 273K.
3. This temperature is called melting point.

Latent Heat of Fusion(L_f )

1. The heat energy required to convert 1gm of solid completely into liquid at a constant temperature is called Latent Heat of fusion.
2. Latent heat of fusion, $$ {L_f=\frac{Q}{m}} $$ 3. Latent heat of fusion is 80 cal/gm.

Units :-

1. SI unit is J/kg.
2. CGS unit is cal/gm.

1. The process in which a substance in liquid phase changes to solid phase by losing some of its energy is called freezing.
2. Freezing of water takes place at 0°C temperature and at one atmospheric pressure.

Water expands on freezing, Floating of ice in water

Aim

To prove water expands on frezzing

Required apparatus :-

Glass bottle, water, refrigerator.

procedure :-

1. Take a small glass bottle with a tight lid.
2. Fill it with water completely without any gaps and fix the lid tightly in such a way that water does not come out of it.
3. Put the bottle into the deep freezer of a refrigerator for a few hours.

observations :-

Take it out from the fridge and you will observe that the glass bottle breaks.

resons :-

1. We know the volume of the water is equal to the volume of the bottle.
2.When the water freezes to ice, the bottle is broken.
3.This means that the volume of the ice should be greater than the volume of the water filled in the bottle.

conclusion :-

We concluded that water ‘expands’ (increases in volume) on freezing.

Floating of ice in water

1.The volume ice is greater than volume of water for the same amount of mass.
2.The density of ice is less than the density of water. So ice floats on water.

Video lessons

Note:- click on synapsis for short notes, videos for video lessons, TQs for taxtual questions and Bits for bits questions

1.Heat part -1

2.Heat part- 2.1

3.Heat part - 2.2 | ఉష్ణం

4. Heat part - 3.1 | ఉష్ణం | మిశ్రమాల పద్ధతి సూత్రం | 10th class telugu medium ps

5.Heat part 3.2 | specific of lead shots | ఉష్ణం సీసపుగుళ్ళ విశిష్టోష్ణం

6. heat part -4 evaporation | ఉష్ణం 4వ భాగం భాష్పీభవనం Science lesson with sanitizer

7. ఉష్ణం part -5 | సాంద్రీకరణం | Heat | condensation

8. ఉష్ణం part -6 |ఎసరు చెప్పే సెన్స్ | మరగడం | boiling

9. Heat part 7 | Science behind melting of ice | కరిగే ఐస్ చెప్పే సైన్సు పాఠం

Heat textual questions

Ans:- Given
m ₁= 50gm; T₁ = 20℃
m₂ = 50gm; T₂ = 40℃
Formula:-
Final temperature, $$ {T=\frac{m_1T_1+m_2T_2}{m_1+m_2}} $$ $$ {T=\frac{50×20+50×40}{50+50}} $$ $${T= \frac{1000+2000}{100}}$$ $${T= \frac{3000}{100}}$$ $${T= 30}$$ ∴ The final temperature of a mixture is 30℃.

Ans:-1. Dogs don't have sweat glands on their skin like we have. Their body is covered with hair. They have sweat glands only in their feet.
2. So It is difficult to evaporate sweat to cool their bodies.
3. panting helps them to decrease body temperature.
4. In this process water present on the tongue evaporates.
5. We know evaporation is a cooling process.
6. That’s why dogs pant during hot summer days to regulate their body temperature.

Ans:- 1. Air contains water in the form of water vapour.
2.The temperature of a cool drink bottle is less than the air surrounding it.
3. As air moves, water vapour in it strikes the surface of the cold soft drink bottle and loses heat.
4. This loss of heat energy decreases their kinetic energy resulting in the change of state from vapour to liquid (water).
5. This is the reason for dew on the surface of a cold soft drink bottle kept in open air

Ans:-

Evaporation	Boiling
1. The process of escaping of molecules from the surface of a liquid at any temperature and pressure is called evaporation.	The process in which the liquid phase changes into gaseous phase at a constant temperature and pressure is called boiling.
2. Evaporation takes place at any temperature and pressure.	Boiling takes place at a constant temperature which is called as boiling point
3.In this process molecules in surface changes its phase from liquid to gas	In this process all the molecules starts changing into liquid to gas
4. It is a surface phenomenon	It is a bulk phenomenon
5.In this process the temperature of the system decreases.	In this process the temperature of the system remains the same.
6. It is a cooling process.	It is a heating process.

Ans: 1. Air contains water molecules in the form of vapour.
2. During condensation water vapour loses its kinetic energy in the form of heat and gets converted into droplets (H₂O).
3. condensation is a warming process.
4. The heat energy lost by the vapour molecules is gained by the surrounding air.
5.The heat energy of the surrounding air rises. So, the surrounding air become warm when vapour phase of H₂O condenses.

Ans:-a) Given:- m = 1gm
Latent heat of vaporisation of water = 540 Cal/gram
Formule:-
Q_L = m×L = 1×540 = 540 Cal

----------------------------------------

b)Given:- m = 1gm
Temperature change from 100℃ to 0℃
so,   ∆t = 100-0 =100℃
m= 1gm ,   s= 1cal/gm ℃
Q2 = ?

Formula:-
Q = mS∆T
Q2 = 1× 1× 100 = 100 Cal

----------------------------------------

Ans:- c) Given:- m = 1gm
Latent heat of fusion is 80 cal/gm.
Formule:-
Q_f = m×L = 1×80 = 80 Cal

----------------------------------------

Ans:-D) Total Heat transferred = Heat transferred to change in state from steam at 100°C to water at 100°C + Heat transferred to cool from 100℃ to 0℃

Part 1) state change from steam at 100°C to water at 100°C
Given :- m = 1gm
Latent heat of vaporisation of water = 540 Cal/gram
Q1 = ?

Formule:-
Q_L = m×L
Q₁=1×540 = 540 Cal

Part 2) Temperature change from 100℃ to 0℃
give m= 1gm,   S= 1 cal/gm ℃
∆t = 100-0 =100℃
Q₂ = ?

Formula:-
Q = mS∆T
Q₂ = 1× 1 × 100 = 100 Cal

∴ Total heat required
Q = Q₁ + Q₂ = 540 + 100 = 640 Cal

------------------------------

Ans:-E) The Heat is released when steam changes into ice.
Total Heat released = Heat released to change in state from steam at 100°C to water at 100°C + Heat released to cool from 100℃ to 0℃ + Heat released to change in state from water at 0°C to ice at 0°C.

Part 1) State change from steam at 100°C to water at 100°C
Given :- m = 1gm ;  Latent heat of vaporisation of water = 540 Cal/gram

Formule Q_L = m×L
Q₁=1×540 = 540 Cal

Part 2) Temperature change from 100℃ to 0℃
Given:- Mass m= 1gm
Specific heat s= 1cal/gm℃
Temperature difference ∆t = 100-0 =100℃
Q₂ = ?

Formula:-
Q = mS∆T
Q₂ = 1× 1× 100 = 100 Cal

Part 3) State change from water at 0°C to ice at 0°C
Given m= 1gm
Latent heat of fusion Q_f= 80cal/ gm
Formule Q_f = m×L
Q₃=1× 80 = 80 Cal

∴ Total heat Released
Q = Q₁ + Q₂ + Q₃
= 540 + 100 + 80 = 720

ans:-
Aim: - To find the specific heat of a given solid.
Materials required: - Calorimeter, thermometer, mixer, stirrer, water, steam chamber, wooden box and lead shots
Procedure: -
1) Measure the calorimeter mass m1 along with the stirrer.
2) Now, fill 1/3 of a calorimeter with water and measure the mass m₂.
3) Only the mass of water becomes m₂ - m₁.
4) measure the temperature of water T₁
in calorimeter with thermometer
5) Take some lead shots, put them in boiling water or steam chamber and heat them up to 100℃ and measure its temperature T₂.
6) To prevent heat loss, the lead should be quickly transferred into a calorimeter.
After a while the mixture reaches a constant temperature.
7) Now measure calorimeter mass m₃, temperature T₃, which contains water and lead.
8) The mass of only lead shots is m₃ - m₂,
9) Let the specific heat calorimeter, solids (lead shots) and water be SL SL and SW respectively.
10) According to the formula of the method of mixtures...
Assuming that the heat Is not lost to surroundings,
The heat lost by the lead shots = the heat absorbed by the calorimeter + the heat absorbed by the water
11) (m₃ - m₂)S_L(T₂ - T₃ ) = (m₁ S_c(T₃ - T₁) + m₁ ) S_w ((T₃ - T₁)
12)(m₃ - m₂)S_L(T₂ - T₃ ) = [m₁ S_c + (m₂ - m₁ ) S_w] (T₃ - T₁)
13)$$ {S_l=\frac{[m_1s_c + (m_2-m_1) S_w](T_3-T_1)}{(m_3-m_2)(T_2-T_3)}} $$
14) If the specific heats of calorimeter and water are known, the specific heat of lead shots(solid) can be calculated with the above equation.

Ans:-

Formula :- 0℃ = 273K

So, 20℃= 20 +273K = 293K

20°C = 293K

Ans:- 1. Which process is called the cooling process ?
2. Which process is called the warming process ?
3. At which temperature evaporation takes place?
4. At which temperature boiling takes place?
5. Do these temperatures are the same?

Ans:-1.The process of escaping of molecules from the surface of a liquid at any temperature and pressure is called evaporation.

2. When wet clothes dry, due to more surface area of clothes, wind speed, and temperature in the surroundings the water present in them becomes vapour and mixes with surrounding air. 3. Evaporation makes wet clothes dry.

Ans:-1.Surface area is one of the factors which affects the rate of evaporation.
2. The surface area of the water in a dish is greater than the surface area of the water in the cap.
3. Here equal amounts of water is taken in both the cup and dish. Hence the water in the dish evaporates quickly.

Ans:- Ans:1. Take a few drops of spirit or petrol(say 1ml) in a test tube and two petri dishes.
2.Now, label the test tube as A and dishes as B and C
3. Keep A and B in a normal room and C in an A.C room for a same period of time
4. After some time if we compare the liquids in A & B, We will observe that liquid evaporates faster in B than A, which contains a larger surface area.
5. So, we can say the rate of evaporation depends on its surface area.
6. When we compare the liquids in B & C, We will observe that liquid evaporates faster in C than B, which is kept in an A.C room due to less humidity (A.C removes water vapour present in that room).
7.So, we can say the rate of evaporation depends on the vapour already present in the surrounding air(humidity).

Ans:-
1. Place a Pyrex glass funnel with its mouth-down in a sauce pan full of water, in such a way that the stem tube of the funnel is above the water or pointing upward into air.
2. Rest the edge of the bottom portion of the funnel on a nail or on a coin so that water can get under it.
3. Place the pan on a stove and heat it till it begins to boil.
4. The boiling point of the water increases with increasing the pressure.
5. So the bubbles first form at the top of the funnel.

Working of Geyser:
1. The Geyser works on the principle of electrical energy converted into heat energy.
2. When heat energy increases, the pressure inside of the Geyser also increases.
3. So, the bubbles of water will come out from the top portion of the Geyser.

Ans:-
1. The sun delivers a large amount of energy to the Earth daily.
2.The water sources on Earth, particularly the oceans, absorb this energy for maintaining a relatively constant temperature.
3.The oceans behave like heat “store houses” for the earth.
4. They can absorb large amounts of heat at the equator without appreciable rise in temperature due to high specific heat of water.
5. Therefore, oceans moderate the surrounding temperature near the equator.
6. Ocean water transports the heat away from the equator to areas closer to the north and south poles.
7.This transported heat helps moderate the climates in parts of the Earth that are far from the equator.
8. This makes the role of high specific heat value of water stabilizing atmospheric temperature more appreciable.

Ans: 1. Watermelons contain a large amount of water.
2. Water specific heat is 1 cal/gm-°C which is greater than any other substance.
3.If the specific heat is high, the rate of rise or fall in temperature is low.
4. so, water in watermelon takes more time to lose or gain heat.
5. Hence specific heat of water plays an important role in keeping a watermelon cool for a long time after removing it from a fridge on a hot day.

Ans:-1.You feel warm when you stay in the bathroom after a shower because of condensation of vapour molecules on your skin.
2.The number of vapour molecules per unit volume in the bathroom is greater than the number of vapour molecules per unit volume outside the bathroom.
3. This excess amount of water vapour tries to condense on you when you try to wipe your body with a towel.
4. Condensation is a warm process. so condensing water vapour molecules on your skin makes us feel warm.

Ans:-Given A at 30°C,
B at 303K =303 - 273 = 30℃,
C at 420k = 420 - 273 = 147.
i)A and B
ii)From C

objective typ questions

Bits and Answers

1. The SI unit of specific heat is_______
2.________ flows from a body at higher temperature to a body at lower temperature.
3.__________ is a cooling process.
4. An object ‘A’at 10 °C and another object ‘B’at 10K are kept in contact, then heat will flow from_________ to_____
5. The latent heat of fusion of ice is ______
6. Temperature of a body is directly proportional to__________
7. According to the principle of method of mixtures, the net heat lost by the hot bodies is equal to ___________by the cold bodies.
8. The sultryness in summer days is due to_________
9. _________is used as a coolant.
10. Ice floats on water because___________

1. Which of the following is a warming process? [   ]
a) Evaporation  b) condensation  c) boiling  d) all the above

2. Melting is a process in which solid phase changes to [   ]
a) liquid phase  b) liquid phase at constant temperature  c) gaseous phase  d) any phase

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

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

5. Specific heat S= [   ]
a) Q/∆t  b) QΔt  c) Q/m∆t  d) m∆t/Q

6. Boiling point of water at normal atmospheric pressure is [   ]
a) 0°C  b) 100°C  c) 110°C  d) -5°C

7. When ice melts, its temperature [   ]
a) remains constant  b) increases   c) decreases  d) cannot say

Answers

1. J/kg-k ,  2. heat,  3. evaporation,
4. A to B,  5. L_f =Q/m,  6. total internal energy,  
7. the net heat gained,  8. humidity,  9. water,
10. density of ice is less than water.

1. a) Evaporation,  2. b) liquid phase at constant temperature,  3. a) 45°C,
4.b) 57°C,   5. c) Q/m∆t,   6. b) 100°C,  7. a) remains constant.

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