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# Change of State, calculate the heat of vaporization of water.

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In an experiment, a certain amount of steam was allowed to mix with a known mass of cold water whose temperature is known. The change in temperature was noted. From the data obtained, calculate the heat of vaporization of water.

Weight of calorimeter = 100.0g
Specific heat of calorimeter = 0.1cal/g°C
Mass of water = 400.0g
Mass of condensed steam = 27.6g
Temperature of cold water = 5.0°C
Temperature of steam = 100.0°C
Final Temperature of mixture = 45.0°C

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RESPONSE: First a bit of basic physics.

Note: A kilogram or a gram is an amount of mass M. The weight, W, of an object is the force exerted by gravity and the relation between the two is: W= M g.

The specific heat, C, of a substance is the amount of heat which, when added to 1 gram of the substance, would increase its temperature by 1 °C (or when removed from 1 gram of it would lower the temp. by 1 degree.)

To obtain the relation between heat Q, added to a mass M of a substance, and the temperature change from To to Tf, we write: (1) Q = M c (To - Tf)

The heat of vaporization Ls of a substance is the amount of heat which changes the state of one gram of it from boiling temperature to steam or from steam changes 1 gram of it to liquid at its boiling temperature.
To obtain the relation between Heat Q which changes a mass M of a substance from steam to water at boiling temperature whose heat of vaporization is Ls, we write: (2) Q = M Ls

In your experiment: By conservation of energy, the heat loss by steam changing state to hot water, plus heat loss by that hot water cooling to final temperature, must equal heat gain by water and by calorimeter each increasing temperature from original to final temperature. Identifying given quantities by letters write: steam Ts= 100 °C and mass Ms= 27.6 g
calorimeter: mass Mc = 100 g specific heat Cc= .1 cal/g °C water mass Mw = 400 g and specific heat Cw= 1 cal/ g °C
water and calorimeter temp. To= 5 °C final temperature of all items: Tf= 45 °C. Using (1) and (2) equations:

Ms Ls + Ms Cw (Ts - Tf) = Mc Cc (Tf - To) + Mw Cw (Tf - To) in which only the heat of vaporization Ls is not given.
The above is the physics part. The arithmetic part, substitution and calculation, should give you.
Ls = 539 cal/g

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 4, 2022, 3:24 pm ad1c9bdddf>
https://brainmass.com/physics/temperature/change-state-calculate-heat-vaporization-water-291941