Chhattisgarh State Board Class 7 Science Chapter 7 Heat and Temperature Exercise Multiple Choice, Matching , Questions and Answers here.
(1) Choose the correct answer –
(1) On heating an object, it’s expansion depends –
(a) On the initial size or the volume of the object
(b) On increasing in temperature
(c) On the material used to make the object
(d) On all the above
Answer:- (d) On all the above.
(2) With 10 calories heat. The temperature of 2g of water will increases by an object depends –
(a) 20c
(b) 50c
(c) 80c
(d) 100c
Answer:- (b) 50c.
(3) The magnitude of heat absorbed by an object depends –
(a) On the mass of the object
(b) On the nature of the object
(c) On the increasing in temperature
(d) On all of them
Answer:- (d) On all of them.
(4) The heat required to increase the temperature of 1g of water by 10c –
(a) 1 calories
(b) 1 kilo calories
(c) 1 Joule
(d) 1 kilo Joule
Answer:- (a) 1 calories.
(5) Out of the following which is not used for measurement of amount of heat –
(a) Calorie
(b) 0c
(c) Kilo calorie
(d) Joule
Answer:- (b) 0c.
(2) Fill in the blanks –
(a) The growth of mammal cells needs a temperature of —- 0c.
Answer:- 37.
(b) Hot water is filled in bottles to massage because it has a high ——–.
Answer:- Specific heat capacity.
(c) The rate of expansion of mercury is almost the same at all temperatures, therefore it is used in a ———.
Answer:- Thermometer.
(d) One calories of heat is equal to ——– joule.
Answer:- 4.186.
(e) The boiling point of water is —— 0c.
Answer:- 100.
(3) Give reason for following –
(a) When boiling water is poured in to a tumbler made up of thick glass, the tumbler cracks.
Answer:- When boiling water is poured in to a tumbler made up of thick glass, it inner surface is expand more than that of outer surface of glass, due to heat of a water so, the glass tumbler cracks.
(b) In summers to keep the water cool we keep it in pitchers made of red soil & not in metal pitchers,
- Pitchers made of red soil has small pores in it wall through which water come on pitches surface.
- Water on surface get evaporate due to air & for change it state it required energy/heat which it take from pitchers water. So there is reduce in temperature of pitcher made up red soil. Metal is good conductor of heat it we used pitcher made up metal it absorbed excess heat of atmosphere result, increasing in temperature of water. Therefore, we used pitcher made up red soil for cool water.
(c) The mercury in the thermometer rises up getting hotter.
Answer:- When the bulb of thermometer is dipped in hot water/or touch hot surfaces, the mercury, in the bulb rises due to thermal expansion.
(d) While sweating contact with air makes us feel coal.
Answer:- While sweating contact with air makes us feel cool because the sweat of our body evaporate due to air. & for the change it state (liquid to air) our body is used. (energy)
(e) When a gas filled balloon is brough near fire, it burst.
Answer:- The thermal expansion of air inside the balloon take place due to fire/heat. So, when a gas filled balloon is brough near fire it burst.
(4) Write two uses of water expansion in daily life.
Answer:- (1) For cooling the engines of motor cars, water in the radiator is not filled up to the top.
(2) At low temperature, water on pond surface frees & cover inner surface by layer of ice, therefore, the aquatic life survive.
(5) On what factor does the heat required by an object to increase its temperature depend?
– The energy to be absorbed by object to increase it’s temperature depend on
(a) Mass of object – large the mass of the object, higher the heat required
(b) Extent of the change in temperature – higher the increasing in temperature, large the energy required.
(c) Nature of material of the object – energy needed depend on the nature of material.
(6) Calculate the heat required for the following –
(a) To increasing in temperature of 0.5 kg water from 250c to 800c
= T1 = 250c M= 0.5kg
T2 =800c
Therefore, we know that, Q = Mass x specific heat capacity x change in temperature
Q = mst
Q = 0.5x4185x(80-25)
= 0.5x4185x55
= 115087.5
= 1.15087×105 Joule
(b) To increasing the temperature of 12 kg copper by 500c
= M= 12 kg, T= 500c
We know that, Q = mst
= 12x386x50
= 231600
= 2.31600×105 Joule
