Test: Calorimetry - NEET MCQ

# Test: Calorimetry - NEET MCQ

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## 10 Questions MCQ Test Physics Class 11 - Test: Calorimetry

Test: Calorimetry for NEET 2024 is part of Physics Class 11 preparation. The Test: Calorimetry questions and answers have been prepared according to the NEET exam syllabus.The Test: Calorimetry MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calorimetry below.
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Test: Calorimetry - Question 1

### Water is used as coolant in automobiles radiators because

Detailed Solution for Test: Calorimetry - Question 1

Water is used as a coolant in automobiles radiators because it has high specific heat capacity. So, it absorbs a large amount of heat for a degree rise in temperature.

Test: Calorimetry - Question 2

### Equal masses of three liquids of specific heats C1, C2 and C3 at temperatures t1, t2 and t3 respectively are mixed. If there is no change of state, the temperature of the mixture is

Detailed Solution for Test: Calorimetry - Question 2

For the composite system, energy conservation yields no net energy flow in or out of the system.
Let final temperature be T
Then, heat absorbed by A+heat absorbed by B+heat absorbed by C=0

Here in the question m1=m2=m3

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Test: Calorimetry - Question 3

### Which of the given relation is true for molar heat capacity of a substance?

Detailed Solution for Test: Calorimetry - Question 3

Test: Calorimetry - Question 4

A piece of iron of mass 100g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.

Detailed Solution for Test: Calorimetry - Question 4

Mass of Iron = 100g
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0ºC
Siron = 470J/kg°C

Total heat gained = Total heat lost.

So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20)
⇒ 47θ – 47 × 60 = 25 × 42 × 40
⇒  θ = 4200 + 2820/47= 44820/47 =953.61°C

Test: Calorimetry - Question 5

A device in which heat measurement can be made is called

Detailed Solution for Test: Calorimetry - Question 5

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

Test: Calorimetry - Question 6

Among the following substances, which one has highest specific heat capacity?

Detailed Solution for Test: Calorimetry - Question 6

Test: Calorimetry - Question 7

The amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2°C at constant pressure is (universal gas constant = R)

Detailed Solution for Test: Calorimetry - Question 7

At constant pressure
dQ= nCpdT
=1*(5R/2)*2
=5R

Test: Calorimetry - Question 8

5 g of ice at 0° C is mixed with 10 g of water at 10° C. The temperature of the mixture is:

Detailed Solution for Test: Calorimetry - Question 8

Heat absorbed by 5g ice when it converted to at 0° C = 5 x 80 = 400 cal.
Heat liberated by 10g water at 10° C to 0° C = 100 cal
Hence there is 15g water at 0° C and 300 cal needs to be liberated , thus for some amount of water converts into ice, hence the temp of mixture is 0° C.

Test: Calorimetry - Question 9

Which of the following relation is true for the specific heat capacity of substance?

Detailed Solution for Test: Calorimetry - Question 9

∆Q =mS∆T

Test: Calorimetry - Question 10

According to law of calorimetry, which of the given relation is true?

Detailed Solution for Test: Calorimetry - Question 10

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat lost by hot body equal to the total heat gained by a cold body.
i.e. heat loss = heat gain

## Physics Class 11

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## Physics Class 11

102 videos|411 docs|121 tests