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# Latent heat and energy during a phase transition

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A 50.0 gram ice cube at -2 oC is heated until all of it is turned to steam at 100.0 oC. How much energy in calories was added to do this?

https://brainmass.com/physics/specific-heat/latent-heat-energy-during-phase-transition-81598

## SOLUTION This solution is FREE courtesy of BrainMass!

Step 1: Convert all temp in to the Kelvin scale or Absolute scale by adding 273
Step 2: Convert all units to SI system ie: 50 gm =0.05Kg.
Consider 50 gm = 0.05 kg of ice at a temperature toC below the normal melting point of ice is converted to the normal melting point of ice = 273 K. Let it is gradually heated, and converted to steam at 100oC or 373K. The total change in energy during the process
which follows

a. The Quantity of heat dQ to be supplied to it so that its temperature if ice rises by dT ( from 271 K to 273 K)

Here dT = 2K= (273K-271K)
C ice =2.1 X 103 J Kg-1 K-1 is the specific heat capacity of ice
M=50 gm =0.05 kg
=0.05 X 2.1 X 103 X 2 =105 J

b. Change in energy or energy absorbed when iceat 273 is converted to water at 273 K
The heat absorbed when unit mass of ice is melt into water is called latent heat l1.(More clearly it is the specific latent heat of fusion of ice). For ice to water conversion l1=3.3 x 105 J Kg-1
Here in this case for 50 gm of ice it is
=0.05 x 3.3 x 105 J = 16500 J

c. Change in energy or energy absorbed when water at 273K is converted to water at 373 K

dT = 100K = (373K-273K)
C water =4185 J Kg-1 K-1 is the specific heat capacity of ice
M=50 gm =0.05 kg
=0.05 x 4185 x 100=20925 J

d. Change in energy or energy absorbed when water at 373K is converted to Steam at 373 K
For that as in the earlier case there will be a parameter called specific latent heat of steam l2=226 x 104 J Kg-1 with this

=0.05 kg X 226 x 104 = 113000 J

The total energy for this process

= 105+16500+20925+113000=150 530 J
In the question the energy is asked in calorie. We obtained it in joule (J). To convert joule into calorie use the conversion given below
1 cal = = 4.1868 J or 1 J = 1/41868 cal = 0.2388459 cal
Then dQ= 150 530 J=150 530 x 0.2388459 = 35953.5 Cal is the required answer.

Note:- Please note that latent heat is the heat absorbed or liberated when matter is converted from one phase to another phase. Eg- water to steam, steam to water, water to ice , ice to water etc. The term specific is stands for, for unit quantity, eg , for one kg. I advise you to check the numerical values with your calculator also.

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