Chhattisgarh State Class 10 Science Chapter 3 Solution

Chhattisgarh State Board Class 10 Science Chapter 3 Heat and Temperature Exercise Multiple Choice, Fill in the Blanks, Questions and Answers here.

Chhattisgarh State Class 10 Science Chapter 3 Solution

1) Choose the correct option-

(i) Ratio of linear, area and volume expansion in solids is-

(a) 1:1:1 (b) 1:2:3 (c) 1:2:1 (d) 3:2:1.

Ans: – (c) 1: 2: 1.

(ii) Amongst object A and B, if the specific heat of object A is less than of object B, then-

(a) Object A will get warm sooner

(b) Object B will get warm sooner

(c) Both will get warm at a same rate

(d) None of above.

Ans: – (a) object A will get warm sooner.

(iii) Which one amongst the following is best conductor of heat-

a) Iron (b) Asbestos (c) Glass (d) Wood.

Ans: – (a) iron

(iv) In which one of the following heats cannot flow due to convection-

a) Tea (b) Water (c) Wind (d) Vacuum.

Ans: – (c) wind.

2) Fill in the blanks-

(i) Heat is not matter instead it is …………. transmitted heat…………….

(ii) ……… Temperature changes……………… results in heat transfer.

(iii) Heat transfer is possible in …. Solid………………… due to conduction.

(iv) …….. Medium…………………. is not required for heat transfer due to radiation.

(v) Heat provided to change the physical state of matter results in change in temperature, thisprocess is called …………… conduction………………………..

3) When two objects with different temperatures are put in contact with each other, they both achieve the same temperature after some time. Why is it so?

Ans: – The main reason for transmission of heat between two object is the temperature difference between them. Heat always transfers from one medium of higher temperature to the other matter which have lower temperature comparing to the other object until attain the same temperature.

4) What is latent heat?

Ans: – When an object changes its state from solid to liquid or liquid to gas it requires a constant amount of heat which is known as latent heat. It unites is Cal/g.

5) Balloon filled with gas bursts when we take it close to fire, why?

Ans: – As we all know when temperature is increases the volume of air is increase. In this case the air tight balloon when came in closer to the fire its volume increases which creates pressure on its surface. As results of that its burst’s.

6) Convert the temperatures given below into given units-

(i) 14° F into Celsius. (ii) 100° C into Fahrenheit. (iii) 12 K into Celsius.

Ans: – As we all know the relations between this scale is,

C/100 = F – 32 / 180 = K – 273 / 100.

By applying this formula, we get

a) Here F = 14°F;

so (14 – 32 )/180 = C/ 100.

Or, C= -100/ 10 = – 10℃.

b) Here C = 100℃;

So, 100/100 = (F – 32) / 180.

Or, F = 212°F.

c) K= 12 k.

So, (12 – 273) / 100 = C / 100.

Or, C = 261℃.

7) Write three uses of specific heat of water in daily life.

Ans: – The specific heat of the water is reason behind its uses in modern life for examples: –

(1) Water is used in hot water bottles because of its specific heat;

(2) For wine preparation water is used and

(3) In the car radiator water is used to cooled because of its specific heat.

8) Determine the final temperature if we mix one kg of water which is at 40°C with one kg of water which is at 60°C.

Ans: – The final temperature of any object is,

T = ( m1 × t1 + m2 + t2) / (m1 + m2).

Here m1 = m2 = 1kg ; t1 = 40℃ and t2 = 60℃.

So , T = (40 + 60) / 2 = 50℃.

9) Mass of an Aluminium object is 500 g, how much amount of heat needs to be supplied in order to increase its temperature by 40°C? Specific heat of Aluminium is 0.09 J/k °C.

Ans: – As we all know that heat (H) = mst;

Where m = mass of the object = 500g=0.5kg;s = specific heat = 0.09 j/k℃; and t = 40℃.

So, H = 0.5 × 0.09 × 40 = 1.8 j.

10) Length of an aluminium wire is 100 cm, what will be the increase in its length when its temperature is increased to 50°C from 30°C? Coefficient of linear expansion for aluminium is 26×10-6/°C.

Ans: – From the relation we know that, L = L’ ( 1+ a × ∆t).

Here L (final length ) =? ; L’ = 100 cm ; a = 26 × 10^-6 /℃ ; ∆t =(50 – 30 )= 20℃ .

L = 100 × ( 1 + 20 × 26 × 10^-6);

L =  100 × 1.00052 ;

L = 100.052 cm.

So, the increase temperature is (100.05 – 100 = 0.052 ℃).

11) Define specific heat capacity.

Ans: – When a unite temperature is increases the minimum heat energy is required. This minimum heat energy is known as the heat capacity of an object.

12) What are the types of heat transfer? Write about them.

Ans: – The heat always is transfer from one medium which is higher temperature to the lower temperature. This process of transfer heat are defined by some ways that are conduction, radiation, convection. In radiation there is no needed any medium but for the other ways a medium is needed.

13) Give few examples of effects of heat seen in our daily life.

Ans: – There are various examples of heat effects that we are seen in our daily life –

  • Our body sweating in hot summer day because of humid temperature condition.
  • For checking of body temperature by mercury thermometer.
  • The water droplets in surface of water bottle become ice when we taken out from refrigerator.
Updated: December 25, 2021 — 4:19 pm

2 Comments

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  1. Q10 solution last line, increase in length aayega instead of increase in temperature. final ans 0.052 cm.

  2. Q.1) i)
    ans:- (b)1:2:3

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