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Published on: 09/10/2019
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.
A metal ball of heat capacity 50J/0C loses 2000 J of heat. By how much will its temperature fall?
2.
What will be the final temperature if 1,68,000 J of heat is absorbed by 2 kg of water at 30°C?
3.
2875 J of heat is required to melt 115 g of lead at its melting point. Calculate the specific latent heat capacity of fusion of lead.
4.
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
5.
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.
6.
Why are burns caused by steam more painful than those caused by boiling water at the same temperature?
7.
What could be the final temperature of a mixture of 100 g of water at 90ºC and 600 g of water at 20ºC?
8.
Define specific latent heat capacity
9.
How does the water kept in an earthen pot remain cool?
1.
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
2.
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
3.
Mass of lead m = 115 g
Heat required Q = 2875 J
Specific latent heat of lead L = ?
We know that Q = m L
L = \(\frac { Q }{ m } =\frac { 2875 }{ 115 } \)
= 25 Jg-1
4.
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
5.
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
6.
(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.
7.
100 g of water = 90o C
600 g of water \(=20^{\circ} \mathrm{C} \times 6=120^{\circ} \mathrm{C}\)
700 g of water total temperature \(=90^{\circ}+120^{\circ} \mathrm{C}=210^{\circ} \mathrm{C}\)
The final temperature of water is \(=210^{\circ} \mathrm{C} / 7=30^{\circ} \mathrm{C}\)
(OR)
Final temperature of a mixture (t) = ?
\(t =\frac{\left(m_{1} c_{1} t_{1}\right)+\left(m_{2} c_{2} t_{2}\right)}{\left(m_{1} c_{1}\right)+\left(m_{2} c_{2}\right)} \)
\(=\frac{(100 \times 4.18 \times 90)+(600 \times 4.18 \times 20)}{(100 \times 4.18)+(600 \times 4.18)} \)
\(=\frac{37620+50160}{418+2508}=\frac{87780}{2926} \)
\(t =30^{\circ} \mathrm{C}\)
8.
Latent heat is the amount of heat energy absorbed or released by a substance during a change in its physical state without any change in its temperature.
9.
The earthen pot is made up of soil. Soil is the bad conductor of heat. So the pot has minute pores. The least amount of water is leaked through the pores. It prevents to heat does not enter into the pot, so the water is remain cool.
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Tamilnadu Stateboard 9th Standard Subjects
Tamilnadu Stateboard Standards