9th Standard Syllabus & Materials
9th Standard
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Published on: 07/12/2019
Heat
Download Tamil Nadu 9th Standard Science question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Science Test1.
Convert the following:
(1) 1000F to 0C
(2) 400C to Fahrenheit (0F)
(3) 350C to Kelvin
(4) 800K to 0C
2.
Give some practical applications of specific latent heat of ice.
3.
How can you experimentally prove that water is a bad conductor of heat? How is it possible to heat water easily while cooking.
4.
Explain convection in daily life.
5.
On a cold day, it is hard to open the lid of a tight container. But when you gently heat the neck you can easily open the lid. why?
On heating Glass expands and lid contracts
On heating lid expands more than the neck and thus slides easily
Neck becomes slippery on heating
Lid of the bottle cannot bear the heat.
6.
Which one of the following scales has lower fixed point at 0°C?
Kelvin scale
Fahrenheit scale
Celsius scale
All of these
7.
Ice does not melt rapidly because of
high specific heat capacity
high latent of fusion
high heat capacity
high latent heat of fusion
8.
The temperature at which a liquid gets converted into its vapour state is called its________
melting point
boiling point
dew point
freezing point
9.
The amount of heat required to raise the temperature through 10C is called_________
thermal energy
calorie
heat capacity
specific heat capacity
10.
What is the minimum possible temperature? Is there also a maximum possible temperature?
11.
Which factor determines the direction of flow of heat from one body to another?
12.
Define one calorie.
13.
What is the other name of heat capacity?
14.
Heat gained by a body depends upon which factors?
15.
A metal ball of heat capacity 50J/0C loses 2000 J of heat. By how much will its temperature fall?
16.
What will be the final temperature if 1,68,000 J of heat is absorbed by 2 kg of water at 30°C?
17.
Calculate the amount of heat required to convert 200g of ice at 00C into water at 00C Specific latent heat of fusion of ice = 336 Jg-1
18.
What is the amount of heat required to raise the temperature of 5 kg of iron from 30°C to 130°C? Specific heat capacity of iron = 483 Jkg-1C-1.
19.
Why are burns caused by steam more painful than those caused by boiling water at the same temperature?
1.
(1) 1000F to 0C
T(0C) = (T(0F)- 32) / 1.8
= (1000F - 32) / 1.8
T(0C) = 37.70C
(2) 400C to Fahrenheit (0F)
T(0F) = (T(0C) x 1.8) + 32
= 400C x 1.8 + 32
= 72 + 32
T(0F) = 1040F
(3) 350C to Kelvin
(TK) = (T0c)+ 273.15
= 35 + 273.15
(TK) = 308.15 K
(4) 800K to 0C
(T0c) = (TK) - 273.15
(T0c) = 80 - 273.15
(T0C) = -193.150C
2.
Specific latent heat of ice is very high (i.e.) 336 J/g.
1. Due to high specific latent heat of ice, snow on mountains do not melt as a whole, but melts gradually into water with the heat of the sun. If the specific latent heat of ice would not have been so high, all the snow would have melted very quickly and there would have been floods in the rivers.
2. All the water in lakes and ponds in cold places do not freeze all at the same time. If freezes slowly and keeps the surrounding moderate.
3. Drinks are cooled more effectively by ice pieces at 00C and not by water at 00C. This is because 1 g of ice takes away 336 J of heat from the drink to melt into water at 00C.
3.
Water has the highest specific heat capacity. So water absorbs a large amount of heat for unit rise in temperature. Thus water is used as a coolant in car radiators and factories to keep engine and other machinery parts cool. So water is a bad conductor of heat. Specific heat capacity of water is 4200J/kgoC. So it is easy to raise the temperature. So it is possible to heat water easily while cooking.
4.
Hot air balloons:
Air molecules at the bottom of the balloon get heated by a heat source and rise. As the warm air rises, cold air is pushed downward and it is also heated. When the hot air is trapped inside the balloon, it rises.
Breezes:
During day time, the air in contact with the land becomes hot and rises. Now the cool air over the surface of the sea replaces it. It is called sea breeze. During night time, air above the sea is warmer. As the warmer air over the surface of the sea rises, cooler air above the land moves towards the sea. It is called land breeze.
Winds:
Air flows from area of high pressure to area of low pressure. The warm air molecules over hot surface rise and create low pressure. So, cooler air with high pressure flows towards low pressure area. This causes wind flow.
Chimneys:
Tall chimneys are kept in kitchen and industrial furnaces. As the hot gases and smoke are lighter, they rise up in the atmosphere.
5.
(b)
On heating lid expands more than the neck and thus slides easily
6.
(c)
Celsius scale
7.
(d)
high latent heat of fusion
8.
(b)
boiling point
9.
(c)
heat capacity
10.
The minimum possible temperature is 0K. There is no limit to maximum temperature.
11.
Temperature.
12.
The amount of heat energy required to raise the temperature of 1 g water by 10C is called one calorie.
13.
Thermal capacity.
14.
1. Mass of the body
2. Change in temperature of the body
3. Nature of the material of the body.
15.
Heat capacity of ball = 50 J0C-1
Heat lost Q = 2000 J
Fall in temperature t = ?
Heat capacity = \(\frac { Q }{ t } \)
50 = \(\frac { 2000 }{ t } \)
t = \(\frac { 2000 }{ 50 } \)
t = 400C
16.
Heat absorbed Q = 1,68,000 J
m = 2kg
Initial temperature t1 = 30°C
Let final temperature t2 = x0C
Rise is temperature \(\triangle \)t = (t2 - t1)
= (x- 30)0C
Specific heat capacity of water C = 4200 J kg-1°C-1
We know that Q = m x c x \(\triangle \)t
1,68,000 = 2 x 4200 x (x - 30)
x - 30 = \(\frac { 1,68,000 }{ 2\times 4200 } \)
\(\)x - 30 = 20
x = 30 + 20
= 50°C
So, the final temperature of water = 50°C
17.
Mass of ice m = 200 g
Specific latent heat of fusion of ice L = 336 Jg-1
Heat required Q = m L
= 200 x 336
Q = 67,200 J
18.
Mass of iron m = 5 kg
Initial temperature t1 = 30°C
Final temperature t2 = 130°C
Rise is temperature \(\triangle \)t2 = (t2 - t1) = 130 - 30 = 100°C
Specific heat capacity of iron c = 483 Jkg-10C-1
Q = m x c x \(\triangle \)t
= 5 x 483 x 100
= 2,41,500 J
19.
(i) When steam hits our skin, it condenses to water and then cools down to the temperature of skin.
(ii) Now, the energy released will be due to latent heat and fall in temperature.
(iii) Whereas when boiling water hits our skin, there is no phase transition but only fall in temperature and the heat transferred to skin will be only due to cooling.
(iv) Also, the loss of energy that is released from steam hitting our skin occurs quickly and in a small localized area, therefore causing damage to our cells.
9th Standard Syllabus & Materials
9th Standard
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards