Chapter 6.4 - Principle of method of mixtures

In the previous section we saw the details about the specific heat capacity. In this section we will see Principle of method of mixtures.

Let us do an experiment:
1. Take 0.2 kg of cold water in a beaker. Measure it's initial temperature . It is found to be 303 K. It is recorded in the table 6.2 below.
2. Take the same quantity (0.2 kg) of hot water in another beaker. Measure it's temperature. It is found to be 323 K. It is also recorded in the table 6.2 below.
3. Pour cold water into the hot water and stir. Measure the resultant temperature. It is found to be 313 K. It is also recorded in the table 6.2:
Table 6.2

Water	Initial temperature (θ1)	Resultant temperature (θ2)	Difference in temperature (θ2 - θ1)	Heat gained/Heat lost
Cold	303 K	313 K	10 K	8400 J
Hot	323 K	323 K	10 K	8400 J

Based on the recordings, we can do some calculations:
4. The 0.2 kg of water which was initially cold, is present in the resultant mixture. That mass of water has gained some heat. Let us calculate the heat energy that it has gained:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (313 - 303) = 10 K 
(iii) Substituting the known values we get: Q = 0.2×4200×10 = 8400 J
5. The 0.2 kg of water which was hot, is present in the resultant mixture. That mass of water has lost some heat. Let us calculate the heat energy that it has lost:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (313 - 323) = -10 K
• The negative sign indicates that, heat energy is given out 
(iii) Substituting the known values we get: Q = 0.2×4200×-10 = -8400 J
6. With the calculated values, we can fill up the table. The completed table is shown below:

Water	Initial temperature (θ1)	Resultant temperature (θ2)	Difference in temperature (θ2 - θ1)	Heat gained/Heat lost
Cold	303 K	313 K	10 K	8400 J
Hot	323 K	323 K	10 K	8400 J

■ From the table, we can see that:
Heat energy lost = heat energy gained
■ When a hot body is in contact with a cold body, heat flows from the hot body to the cold body, till both the bodies attain the same temperature. The heat lost by the hot body = heat gained by the cold body. This is the principle of method of mixtures

Solved example 6.8
4 kg of hot water at 343 K is added to 6 kg of water at 293 K. The resultant temperature is 313 K. Calculate the heat lost by the hot water and the heat gained by the cold water. Assume that there is no heat loss to the surroundings
Solution:
1. 4 kg of hot water:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (313 - 343) = -30 K
• The negative sign indicates that, heat energy is given out 
(iii) Substituting the known values we get: Q = 4×4200×-30 = -504000 J
2. 6 kg of water:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (313 - 293) = 20 K 

(iii) Substituting the known values we get: Q = 6×4200×20 = 504000 J
3. So we can write:
Heat energy lost by the hot water = heat energy gained by the cold water

Solved example 6.9
In a bucket, there is 8 kg of water at 298 K. To this, 2 kg of water at 353 K is added. Calculate the resultant temperature assuming that the bucket and the surroundings do not receive any heat.
Solution:
1. 8 kg of water:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (θ₂ - 298)
• θ₂ is the resultant temperature. It is the unknown quantity. 
(iii) Substituting the known values we get: Q = 8×4200×(θ₂ - 298)
2. 2 kg of water:
(i) We have : Q = mcθ.
(ii) θ = (θ₁ - θ₂) = (353 - θ₂)
• θ₂ is subtracted from 353 because we know that the 2 kg water will lose heat, and so θ₂ will be lesser than 353
(iii) Substituting the known values we get: Q = 2×4200×(353-θ₂)
3. Heat energy lost by the hot water = heat energy gained by the cold water
So we can write:
8×4200×(θ₂ - 298) = 2×4200×(353- θ₂)
⟹ 8×(θ₂ - 298) = 2×(353- θ₂) ⟹ 4×(θ₂ - 298) = (353- θ₂)  
⟹ 4×θ₂ - 4×298 = 353- θ₂ ⟹ 5 × θ₂  = 1192 +353 = 1545 ⟹ θ₂ = ¹⁵⁴⁵⁄₅ = 309 K

Solved example 6.10
How much water at 370 K is required to raise the temperature of 5 kg water at 300 K to 350 K 
Solution:
• 5 kg water at 300 K is placed in a vessel
• To that, 'm' kg water at 370 K is to be added.
• The resultant temperature must be 350 K. So θ₂ = 350 K
• We have to find 'm'   
1. 5 kg of water:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (350 - 300) = 50  
(iii) Substituting the known values we get: Q = 5×4200×50
2. 'm' kg of water:
(i) We have : Q = mcθ.
(ii) θ = (θ₂ - θ₁) = (370 - 350) = 20  

(iii) Substituting the known values we get: Q = m×4200×20
3. Heat energy lost by the hot water = heat energy gained by the cold water

So we can write:
5×4200×50 = m×4200×20 
⟹ 250 = 20m ⟹ m = ²⁵⁰⁄₂₀  = 12.5 kg

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In the next section, we will see Change of state.

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