1. Definition of Black principle
If we mix hot and cold water in an open container (for example a bucket), heat transfer from hot water to cold water. Because the container is open, then some heat moves into the air. The container also becomes warmer. Heat released by hot water is not only absorbed by cold water, but also absorbed by atmosphere and containers. In this case, the bucket is a non-isolated system. When we mix hot and cold water in a closed thermos, heat transfer from hot water to cold water. Thermos are isolated systems, so no heat moves into the air or a thermos. Cold water only absorbs the heat released by hot water until the mixture of hot water and cold water reaches the thermal equilibrium.
The Black Principle states that in an isolated closed system, heat released by a high-temperature object = heat absorbed by a low-temperature object.
2. Black principle formula
Q release = Q absorbs
Q release = heat released by high-temperature objects, Q absorbs = heat absorbed by low-temperature objects.
3. Examples of questions and discussions
Example of problem 1.
A hot iron with a mass of 1 kg and a temperature of 100 oC is put into a container containing 2 kg of water and a temperature of 20 oC. What is the temperature of the final mixture? The specific heat of iron = 450 J/kg Co, the specific heat of water = 4200 J/kg Co.
Known :
Mass of iron (m) = 1 kg
Temperature of iron (T) = 100 oC
Mass of water (m) = 2 kg
Temperature of water (T) = 20 oC
Specific heat for iron (c) = 450 J/kg Co
Specific heat for water (c) = 4200 J/kg Co
Wanted : T ?
Solution :
Iron has a higher temperature than water so that iron releases heat, water absorbs heat.
Q release = Q absorbs
m c ΔT = m c ΔT
(1)(450)(100-T) = (2)(4200)(T-20)
(450)(100-T) = (8400)(T-20)
45000 – 450T = 8400 T – 168000
45000 + 168000 = 8400 T + 450 T
213000 = 8850 T
T = 213000 : 8850
T = 24 oC
The final temperature of the mixture of hot iron and cold water when both are in thermal equilibrium is 24 oC.
Example of issue 2.
0.2 kg mass of ice mixed with warm tea with a mass of 0.2 kg. Temperature of ice = -10 oC, temperature of warm tea = 40 oC. The specific heat of ice = 2100 J/kg Co, the specific heat of water = 4200 J/kg Co, the heat of fusion for water = 334000 J/kg. Ice and tea are mixed in an isolated, closed system.
Known :
Mass of ice (m) = 0.2 kg
Mass of tea (m) = 0.2 kg
Specific heat of water (c) = 4180 J/kg Co
Specific heat of ice (c) = 2100 J/kg Co
Heat of fusion for water (LF) = 334 x 103 J/kg
Temperature of ice (Tice) = -10 oC
Temperature of tea (Ttea) = 40 oC
Wanted : T mixture
Solution :
First step, estimate the final state
The heat that must be released by water to reduce the temperature of 0.2 kg of warm tea, from 40 oC to 0 oC
Q release = m c ∆T
Q release = (0.2 kg) (4180 J/Kg Co) (40 oC – 0 oC)
Q release = (0.2 kg) (4180 J/Kg Co) (40 oC)
Q release = 33,440 Joule
Q release = 33.44 kJ
Heat is absorbed by 0.2 kg of ice to increase the temperature from -10 oC to 0 oC
Q absorbed = m c ∆T
Q absorbed = (0.2 kg) (2100 J/Kg Co) (0 oC – (-10 oC))
Q absorbed = (0.2 kg) (2100 J/Kg Co) (10 oC)
Q absorbed = 4,200 Joule
Q absorbed = 4.2 kJ
The heat absorbed to melting 0.2 kg of ice (heat needed to melting all ice into water)
Q fusion = m LF
Q fusion = (0.2 kg) (334 x 103 J/kg)
Q fusion = 66.8 x 103 Joule
Q fusion = 66.8 kJ
Based on the above calculations, the results :
Q release = 33.44 kJ
Q absorb = 4.2 kJ
Q fusion = 66.8 kJ
When tea released heat as much as 33.44 kJ, the temperature of tea changed from 40 oC to 0 oC. Some heat released (about 4.2 kJ) is used to raise the ice temperature from -10 oC to 0 oC. 33.44 kJ – 4.2 kJ = 29.24 kJ. The remaining heat = 29.24 kJ. To melt all the ice into water requires a heat of 66.8 kJ. The remaining heat is only 29.24 kJ.
In conclusion, the heat released by warm tea is used only to raise the ice temperature from -10 oC to 0 oC and melting some ice. Some ice has melted into water; some have not. During the process of changing solid into a liquid, the temperature does not change. Therefore, the temperature of the final mixture = 0 oC.
Example of problem 3.
mass of hot tea (m) = 0.4 kg,
mass of ice (m)= 0.2 kg,
temperature of ice = -10 oC
temperature of hot tea = 90 oC.
If both are mixed, what is the temperature of the final mixture?
The specific heat for water (c) = 4180 J/kg Co,
The specific heat for ice (c) = 2100 J/kg Co,
Heat of fusion for water (LF) = 334 x 103 J/kg
Solution :
First step: Estimate the final state
The heat that must be released by water to reduce the temperature of 0.4 kg of hot tea, from 90 oC to 0 oC
Q release = m c ∆T
Q release = (0.4 kg) (4180 J/Kg Co) (90 oC – 0 oC)
Q release = (0.4 kg) (4180 J/Kg Co) (90 oC)
Q release = 150,480 Joule
Q release = 150.48 kJ
The heat absorbed by 0.2 kg of ice to increase its temperature from -10 oC to 0 oC
Q absorbed = m c ∆T
Q absorbed = (0.2 kg) (2100 J/Kg Co) (0 oC – (-10 oC))
Q absorbed = (0.2 kg) (2100 J/Kg Co) (10 oC)
Q absorbed = 4,200 Joule
Q absorbed = 4.2 kJ
The heat needed to melt 0.2 kg of ice (heat needed to turn all ice into water)
Q fusion = m LF
Q fusion = (0.2 kg) (334 x 103 J/Kg)
Q fusion = 66.8 x 103 Joule
Q fusion = 66.8 kJ
Based on the above calculations, the results:
Q release = 150.48 kJ
Q absorbed = 4.2 kJ
Q fusion = 66.8 kJ
When hot tea released 150.48 kJ of heat, the temperature of the hot tea changed from 90 oC to 0 oC. Some heat released (about 4.2 kJ) is used to raise the ice temperature from -10 oC to 0 oC.
150.48 kJ – 4.2 kJ = 146.28 kJ. The remaining heat = 146.28 kJ.
The heat needed to melt all ice into water is only 66.8 kJ. 146.28 kJ – 66.8 kJ = 79.48 kJ. It turned out that the excess was 79.48 kJ. Hot tea does not release all heat until the temperature is reduced to 0 oC. Conclusion: the temperature of the final mixture must be greater than 0 oC.
Step Two: Determine the final temperature (T)
The heat needed to increase the ice temperature from -10 oC to 0 oC = 4200 Joule
The heat had to melt all ice into water aka latent heat = 66,800 Joule
The heat had to increase the temperature of the water (water from all the ice fusion) from 0 oC to T = (ice mass) (heat of water) (T – 0 oC) = (0.2 kg) (4180 J/kg Co)(T) = (836 T) J/Co
The heat released by warm tea to reduce the temperature from 90 oC to T = (mass of hot water) (heat type of water) (90 oC – T) = (0.4 kg) (4180 J/kg Co) (90 oC – T) = 1672 J/Co (90 oC – T) = 150,480 J – (1672 T) J/Co
4200 J + 66,800 J + (836 T) J/Co = 150,480 J – (1672 T) J/Co
71,000 J + (836 T) J/Co = 150,480 J – (1672 T) J/Co
(836 T) J/Co + (1672 T) J/Co = 150,480 J – 71,000 J
(2508 T) J/Co = 79,480 J
T = 31.7 oC
Final temperature = 31.7 oC
