Heat is also energy, we can see that it is moving / transferring from a higher temperature to lower temperature.
2 system let say water and ice, if they are mixed together, what will happen?
Water will be colder while ice will be hotter.
They will stay at the same point (temperature), so it means heat loss from the water will be gained by ice, to make them stay balance in one temperature.
In that case we will have equation
Q loss = Q gain
What happen to solid matter when they are mixed to liquid? or solid matter when mixed to gas? and so on.
In this case, you are going to need a heating graph to see how the heat received will move point to point.
To understand the heating (cooling) graph, observe the picture below
As solid (the yellow one) receives heat, the molecule inside will vibrate faster creating space to expand and increasing the temperature. The same thing also happen to the liquid (the light blue and grey one).
During fusion / melting of solid (the green one), we will not find any temperature changes, only state (phase) changing occur, the absorbed heat will be used to overcome the intermolecular bond (which lead to state changing). This also happen during boiling / vaporizing (the blue one)
Once, you have known the step by step imminence of a substance when losing or gaining heat from the heating (cooling graph), it will be easier for you to calculate heat needed or extracted in a system which contain 2 or more state.
Ex : ice at -5C will be mixed with water at 30C, what will happen?
The ice is gaining heat while the water is losing heat, the final temperature of the two system could be below freezing point, at freezing point or above freezing point, depends on the mass of each ice or water. When the final temperature below freezing point( let say, A), it means the ice will gain heat from its heat capacity to a certain temperature (A) while water will lose heat from 30C to 0C (according to its heat capacity) and then water will change state into ice then followed by the changing temperature of the freezing water to a same number as the ice (A).
When the final temperature at freezing point ( 0C ), it means the ice will gain heat from its heat capacity to 0C then followed by the fusion of the ice into liquid while water will lose heat from 30C to 0C (according to its heat capacity).
When the final temperature above freezing point (let say, B), it means the ice will gain heat from its heat capacity to 0C, followed by fusion, turn to water then rising up its temperature to B (of course, at this poinr, it has turned to water) while water will lose heat from 30C to B.
Worked example :
Q 1 kg of alcohol (heat capacity of 3000 J/kgC) at 40C will be mixed with 1 kg of water (heat capacity of 4000 J/kgC) at 75C. What will be the final temperature of the mixture?
A Alcohol will gain heat from water. Take the final temperature as T then we can picture that the heat gained from the water will be 1 kg x 4000 J/kgC x (75-T)C. On the other hand the water will lose heat as 1 kg x 3000 J/kgC x (T-40). The number of heat gained and loss are the same, so we will get equation : 4000 (75 - T) = 3000 (T - 40) then we can find that the final temperature is 60C
Q 1 kg of ice (heat capacity of 2000 J/kgC and latent heat of fusion 300000J/kg) at -4C will be mixed with 1 kg of water (heat capacity of 4000 J/kgC) at 80C. What will the final temperature of the mixture?
A Ice will gain heat from water. Take the final temperature as 0C (stay at water) then ice will gain heat to raise temperature to 0C as 1 kg x 2000 J/kgC x ((0-(-4))C then followed by melting as 1 kg x 300000 J/kg. On the other hand the water will lose heat as 1 kg x 4000 J/kgC x (80-0). The number of heat gained and loss should be the same, however we get equation that 8000 J + 300000 J = 320000 J which is impossible because the heat loss from the water (320000 J) is over the heat gained by ice, so it means after the ice melting, it will raise temperature, let say A. It will make us to recalculate the heat gained by ice as 8000 J + 300000 J + 1 kg x 4000 J/kgC x A C and the heat loss of the water as 1 kg x 4000 J/kgC x (80-A) C. That end the equation as 308000 J + 4000A = 320000 - 4000 A which lead that the final temperature (A) is 1.5C
Q 1 kg of ice (heat capacity of 2000 J/kgC and latent heat of fusion 300000J/kg) at -4C will be mixed with 1 kg of water (heat capacity of 4000 J/kgC) at 10C. What will the final temperature of the mixture?
A Ice will gain heat from water. Take the final temperature as 0C (stay at water) then ice will gain heat to raise temperature to 0C as 1 kg x 2000 J/kgC x ((0-(-4))C then followed by melting as 1 kg x 300000 J/kg. On the other hand the water will lose heat as 1 kg x 4000 J/kgC x (10-0). The number of heat gained and loss should be the same, however we get equation that 8000 J + 300000 J = 40000 J (which is impossible), from the number we can see that heat needed for ice to become water (308000 J) is higher than the given heat from the water (40000 J) , so we can predict that not all ice become water (only port of ice, let say m) it means the equation will become 8000 J + 300000m = 40000 J, and this equation is possible. In this case the final temperature is 0C
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