मराठी

Fill in the Following Blanks Using Suitable Word: the Amount of Heat Required to Change the State of a Physical Substance Without Any Change of Temperature is Called .......... of the Substance. - Physics

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प्रश्न

Fill in the following blank using suitable word:

 The amount of heat required to change the state of a physical substance without any change of temperature is called .......... of the substance.

रिकाम्या जागा भरा

उत्तर

 The amount of heat required to change the state of a physical substance without any change of temperature is called Latent heat of the substance.

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पाठ 5: Heat - Exercise 4 [पृष्ठ २४८]

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फ्रँक Physics - Part 2 [English] Class 10 ICSE
पाठ 5 Heat
Exercise 4 | Q 5.7 | पृष्ठ २४८

संबंधित प्रश्‍न

A piece of iron of mass 2.0 kg has a heat capacity of 966 J K-1. Find heat energy needed to warm it by 15°C.


Describe an experiment to show that there is absorption of heat energy when the ice melts.


A bucket contains 8 kg of water at 250 C. 2 kg of water at 800 C is poured into it. Neglecting the heat energy absorbed by the bucket, calculate the final temperature of water.

What mass of a liquid A of specific heat capacity 0.84 J K-1 and at a temperature 400C must be mixed with 100 g of a liquid B of specific heat capacity 2.1 J g-1 K-1 and at 200C, so that final temperature of mixture becomes 320C?

Give scientific reasons for the following:

Bottled drinks are cooled more effectively when surrounded by lumps of ice than by cold water at 0°.


100 g of ice at -10°C is heated. It is converted into steam. Calculate the quantity of heat which it has consumed. (Sp. heat of ice = 2100J/kg°C, sp. heat of water= 4200 J/kgK, sp. heat of water = 42000 J/kgK, sp. latent heat of ice = 2260000 J/kg).


A refrigerator converts 100 g of water at 20°C to ice at -10°C in 73.5 minutes. Calculate the average rate of heat extraction from water in watt. (Sp. heat of ice= 2100 J/kgK, sp. heat of water = 4200 J/kgK, Sp. latent heat of ice = 336000 J/kg.)


Fill in the following blank using suitable word:

1 cal = .......... J


Ice is more effective in cooling than the ice-water. Explain.


200 g of hot water at 80°C is added to 300 g of cold water at 10 °C. Calculate the final temperature of the mixture of water. Consider the heat taken by the container to be negligible. [specific heat capacity of water is 4200 J kg-1 °C-1]


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