Thermal Properties of Matter Test

Result

Thermal Properties of Matter Test - 7

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Water is used as coolant in automobiles radiators because

A
It has high specific heat capacity

B
It is easily available

C
It is easy to carry

D
It is cheap

Solution

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.
Question 2
1 / -0

Equal masses of three liquids of specific heats C₁, C₂ and C₃ at temperatures t₁, t₂and t₃ respectively are mixed. If there is no change of state, the temperature of the mixture is

A
(C₁t₁ + C₂t₂ + C₃t₃)/3(C₁ + C₂ + C₁)

B
(C₁t₁ + C₂t₂ + C₃t₃)/(C₁ + C₂ + C₁)

C
(t₁ + t₂ +t₃ )/3

D
3(C₁t₁ + C₂t₂ + C₃t₃)/(C₁ + C₂ + C₁)

Solution

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
Question 3
1 / -0

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

A

B

C

D
Question 4
1 / -0

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.

A
500°C

B
900°C

C
953.6^∘C

D
706.80 °C

Solution

Mass of Iron = 100g
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0^ºC
S_iron = 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
Question 5
1 / -0

A device in which heat measurement can be made is called

A
Joule meter

B
Calorimeter

C
Thermal meter

D
Gauge meter

Solution

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.
Question 6
1 / -0

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

A
kerosene

B
Water

C
Edible oil

D
Ice

Solution
Question 7
1 / -0

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)

A
5R/2

B
3 R

C
2R

D
5 R

Solution

At constant pressure
dQ= nCpdT
=1*(5R/2)*2
=5R
Question 8
1 / -0

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

A
2°C

B
0°C

C
5°C

D
2.5°C

Solution

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.
Question 9
1 / -0

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

A

B

C

D

Solution

∆Q =mS∆T
Question 10
1 / -0

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

A
Heat gained ≥ Heat lost

B
Heat gained = Heat lost

C
Heat gained > Heat lost

D
Heat lost > Heat gained

Solution

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