Heat Exchange in Mixtures & Thermodynamics

[gomani]

i would like to know more about the heat exchange in mixtures and thermodynamics
example would be like:

How much heat energy is required to change?
a) 2.0kg of ice at 0 C to water at 0 C?

b) 500g of water at 100 C to steam at 100 C?

 

[Andrew Mason]

i would like to know more about the heat exchange in mixtures and thermodynamics
example would be like:

How much heat energy is required to change?
a) 2.0kg of ice at 0 C to water at 0 C?

b) 500g of water at 100 C to steam at 100 C?

These change of state transitions are the result of molecules having sufficient energy to overcome bonds with other water molecules. This bond-breaking heat energy does not increase temperature (ie molecular kinetic energy) and is called latent heat (of fusion / vaporization). Just look up these latent heats to find the heat energy involved in each of these examples.

AM

 

[kyle8921]

Thank you for your interest in heat exchange in mixtures and thermodynamics. To answer your question, the amount of heat energy required to change the state of a substance is determined by its specific heat capacity and the heat of fusion or vaporization. The specific heat capacity is the amount of heat energy required to raise the temperature of 1kg of a substance by 1 degree Celsius, while the heat of fusion or vaporization is the amount of heat energy required to change the state of a substance without changing its temperature.

a) To change 2.0kg of ice at 0°C to water at 0°C, we would need to provide the ice with enough heat energy to increase its temperature from 0°C to its melting point of 0°C and then to change its state from a solid to a liquid. The specific heat capacity of ice is 2.09 J/g°C and the heat of fusion for ice is 334 J/g. Therefore, the total amount of heat energy required would be:

Q = (2.0kg)(2.09 J/g°C)(0°C) + (2.0kg)(334 J/g) = 668 J + 668 J = 1336 J

b) Similarly, to change 500g of water at 100°C to steam at 100°C, we would need to provide the water with enough heat energy to increase its temperature from 100°C to its boiling point of 100°C and then to change its state from a liquid to a gas. The specific heat capacity of water is 4.18 J/g°C and the heat of vaporization for water is 2260 J/g. Therefore, the total amount of heat energy required would be:

Q = (500g)(4.18 J/g°C)(100°C) + (500g)(2260 J/g) = 20900 J + 1130000 J = 1150900 J

I hope this helps to further your understanding of heat exchange in mixtures and thermodynamics. If you have any other questions, please don't hesitate to ask.