NEET Physics Solutions For Class 11 Chapter 1 Calorimetry

NEET Physics Class 11 Chapter 1 Calorimetry

The branch of thermodynamics which deals with the measurement of heat is called calorimetry.

• A simple calorimeter is a vessel generally made of copper with a stirrer of the same material. The vessel is kept in a wooden box to isolate it thermally from the surroundings. A thermometer is used to measure the temperature of the contents of the calorimeter.
• Objects at different temperatures are made to come in contact with each other in the calorimeter.
• As a result, heat is exchanged between the object as well as with the calorimeter. Neglecting any heat exchange with the surrounding, the total height is conserved i.e. heat given by one is equal to that taken by the other.

Law of Mixture:

When two substances at different temperatures are mixed together, the exchange of heat continues to take place till their temperatures become equal. This temperature is then called the final temperature of the mixture. Here, Heat taken by one substance = Heat given by another substance.

⇒ m₁s₁(T₁− T_m) = m₂ s₂(T_m− T₂)

Question 1. An iron block of mass 2 kg, falls from a height of 10 m. After colliding with the ground it loses 25% energy to its surroundings. Then find the temperature rise of the block. (Take sp. heat of iron 470 J/kg ºC)

Solution:

⇒ \(\mathrm{ms} \Delta \theta=\frac{4}{4} \mathrm{mgh}\)

= \(\Delta \theta=\frac{3 \times 10 \times 10}{4 \times 470}=0.159^{\circ} \mathrm{C}\)

Note: If specific heat is given as calorie / kgºC, it is to be converted to J/kgºC by multiplying with 4.18

Zeroth law of thermodynamics:

If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are also in thermal equilibrium with each other.

Question 2. The temperature of equal masses of three different liquids A, B, and C are 10ºC 15ºC, and 20ºC respectively. The temperature when A and B are mixed is 13ºC and when B and C are mixed, it is 16ºC. What will be the temperature when A and C are mixed?

Solution:

when A and B are mixed

ms₁ × (13 – 10) = m × s₂× (15 – 13)

3s₁= 2s₂….. (1)

when B and C are mixed

s_s × 1 = s_s× 4 …… (2)

when C and A are mixed

s₁(θ – 10) = s₃× (20 – θ) ….(3)

by using equation (1), (2) and (3)

we get θ = \(\frac{140}{11}^{\circ} \mathrm{C} \)

Question 3. If three different liquids of different masses specific heat and temperature are mixed with each other then what is the temperature mixture at thermal equilibrium?

1. m₁, s₁, T₁ → specification for liquid
2. m₂, s₂, T₂ → specification for liquid.
3. m₃, s₃, T₃ → specification for liquid.

Solution:

Total heat lost or gained by all substances is equal to zero

ΔQ = 0

m₁s₁(T – T₁) + m₂s₂(T – T₂) + m₃s₃(T – T₃) = 0

So, T = \(\frac{m_1 s_1 T_1+m_2 s_2 T_2+m_3 s_3 T_3}{m_1 s_1+m_2 s_2+m_3 s_3}\)

Question 4. The following equation calculates the value of H 1kg ice at –20ºC = H + 1 Kg water at 100ºC, here H means heat required to change the state of a substance.

Solution:

The heat required to convert 1 kg ice at – 20ºC into 1 kg water at 100ºC

ΔQ (Kcal) = 1 kg ice at – 20ºC to 1 kg ice at 0ºC ice + 1 kg water

at 0ºC + 1 kg water at 0ºC to 1 kg water at 100ºC

= 1 ×\(\frac{1}{2}\)× 20 + 1 × 80 + 1 × 100 = 190 Kcal.

So H = – 190 Kcal

The negative sign indicates that 190 Kcal heat is withdrawn from 1 kg water at 100ºC to convert it into 1 kg ice at – 20ºC

Question 5. 1 kg ice at –20ºC is mixed with 1 kg steam at 200ºC. Then find the equilibrium temperature and mixture content.

Solution:

Let equilibrium temperature is 100 ºC heat required to convert 1 kg ice at –20ºC to 1 kg water at 100ºC is equal to

H₁= 1 ×\(\frac{1}{2}\)× 20 + 1 × 80 + 1 × 1 × 100 = 190 Kcal

The heat released by steam to convert 1 kg steam at 200ºC to 1 kg water at 100ºC is equal to

H₂= 1 ×\(\frac{1}{2}\)× 100 + 1 × 540 = 590 Kcal

1 kg ice at – 20ºC = H₁+ 1kg water at 100ºC …… (1)

1 kg steam at 200ºC = H₂ + 1kg water at 100ºC ……. (2)

by adding equation (1) and (2)

1 kg ice at –20ºC + 1 kg steam at 200ºC = H1+ H2+ 2 kg water at 100ºC.

Here heat required to ice is less than the heat supplied by steam so the mixture equilibrium temperature is 100ºC then steam is not completely converted into water.

So the mixture has water and steam which is possible only at 100ºC

mass of steam which converted into water is equal to

m = \(\frac{190-1 \times \frac{1}{2} \times 100}{540}=\frac{7}{27} \mathrm{~kg}\)

so mixture content mass of steam = \(1-\frac{7}{27}=\frac{20}{27} \mathrm{~kg}\)

mass of water = \(1+\frac{7}{27}=\frac{34}{27} \mathrm{~kg}\)