In this week\'s experiment, the heat of vaporization of liquid nitrogen is deter

Question

In this week's experiment, the heat of vaporization of liquid nitrogen is determined by measuring the change in temperature of a known sample of warm water when liquid nitrogen is added.

In one experiment, the mass of water is 106 grams, the initial temperature of the water is 59.3oC, the mass of liquid nitrogen added to the water is 61.3 grams, and the final temperature of the water, after the liquid nitrogen has vaporized, is 32.3oC.

Specific heat of water = 4.184 J K-1g-1

How much heat is lost by the warm water?

Heat lost =  J

What is the heat of vaporization of nitrogen in J g-1?

Heat of vaporization =  J g-1

What is the molar heat of vaporization of nitrogen?

Molar heat of vaporization =  J mol-1

Trouton's constant is the ratio of the enthalpy (heat) of vaporization of a substance to its boiling point (in K). The constant is actually equal to the entropy change for the vaporization process and is most often a measure of the entropy in the liquid state. The value of the constant usually lies within the range 70 to 90 J K-1mol-1, with a value toward the lower end indicating high entropy in the liquid state.

The normal boiling point of liquid nitrogen is -196oC. Based upon your results above, what is the value of Trouton's constant?

Trouton's constant =  J K-1mol-1

Explanation / Answer

The mass of water is 106 grams, the initial temperature of the water is 59.3oC, the mass of liquid nitrogen added to the water is 61.3 grams, and the final temperature of the water, after the liquid nitrogen has vaporized, is 32.3oC.

Specific heat of water = 4.184 J K-1g-1


Solution :

How much heat is lost by the warm water?
Heat loss :

Heat is lost as result of evaporation of Nitrogen from water and that is taken from water.

q = m C delta T

m = mass in g , C is specific heat, Delta T = Change in T

Lets plug value for water

=106 g * 4.184 J /g per deg C * (32.3

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