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Published on: 21/09/2019
Term 2 - 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.
What is the minimum possible temperature? Is there also a maximum possible temperature?
2.
Who introduced the term latent heat?
3.
What is the relation between calorie and Joule?
4.
Define one calorie.
5.
What is the other name of heat capacity?
6.
Define latent heat of fusion?
7.
What do you mean by thermal equilibrium?
8.
Heat gained by a body depends upon which factors?
9.
How much heat energy is required to change 2 kg of ice at 0ºC into water at 20ºC? (Specific latent heat of fusion of water = 3,34,000 J/kg, Specific heat capacity of water = 4200JKg-1K-1).
10.
What is the heat in joules required to raise the temperature of 25 grams of water from 0ºC to 100ºC? What is the heat in Calories?
(Specific heat of water = 4.18 J/gºC)
1.
The minimum possible temperature is 0K. There is no limit to maximum temperature.
2.
Joseph Black in 1750.
3.
1 Calorie = 4.186 J
= 4.2 J
4.
The amount of heat energy required to raise the temperature of 1 g water by 10C is called one calorie.
5.
Thermal capacity.
6.
Heat energy is absorbed by a solid during melting and an equal amount of heat energy is liberated by the liquid during freezing, without any temperature change. It is called latent heat of fusion.
7.
When two bodies at different temperatures are kept in contact with each other, then heat energy flows from the hot body to the colder one.
It means that the hot body will lose heat and the cold body will gain heat till they reach a common temperature. This state is called thermal equilibrium.
8.
1. Mass of the body
2. Change in temperature of the body
3. Nature of the material of the body.
9.
Mass (m) = 2 kg
Specific heat capacity (c) = 4200 Jkg-1 K-1
Temperature variation (\(\triangle \)t) = 20oC -0oC = 20oC
Specific latent heat (L) = 3,34,000 J/kg
Heat energy required = mc\(\triangle \)t + ml
= (2 x 4200 x 20) + (2 x 3, 34,000)
= 168000 + 668000
= 836000 J
10.
Mass of water = 25 grams
\(\triangle\)T = (100oC - 0oC) =100oC
C = 4.18 J/goC
Heat energy required Q = M x C x \(\triangle\)T
= 25 x 4.18 x 100oC
= 10450 J
9th Standard Syllabus & Materials
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards
