Latent Heat and Sensible Heatheat is a form of energy associated with the motion of atoms, molecules and other particles which comprise matter. Heat can be created by chemical reactions (such as burning), nuclear reactions (such as fusion taking place inside the Sun), electromagnetic dissipation (as in electric stoves), or mechanical dissipation (such as friction). Heat can be transferred between objects by radiation, conduction and convection.
Sensible heat When an object is heated, its temperature rises as heat is added. The increase in heat is called sensible heat. Similarly, when heat is removed from an object and its temperature falls, the heat removed is also called sensible heat. Heat that causes a change in temperature in an object is called sensible heat. Latent heat All pure substances in nature are able to change their state. Solids can become liquids (ice to water) and liquids can become gases (water to vapor) but changes such as these require the addition or removal of heat. The heat that causes these changes is called latent heat. Latent heat however, does not affect the temperature of a substance - for example, water remains at 100°C while boiling. The heat added to keep the water boiling is latent heat. Heat that causes a change of state with no change in temperature is called latent heat.
Latent Heat of Fusion To raise the temperature of 1 pound of ice from 0°F to 32°F, you must add 16 BTU. To change the pound of ice at 32°F to a pound of water at 32°F, you add 144 BTU (latent heat of fusion). There is no change in temperature while the ice is melting. Example Five pounds of ice at 32oF is inserted into 25 pounds of water at 68oF. The ice melts and the final temperature of water is 38oF. Find the latent heat of fusion of ice. Solution H1 = latent heat of fusion of ice = heat gained by 5 pounds of ice changing state from solid at 32oF to liquid at 32oF H2 = heat lost by 25 pounds of water cooling from 68oF to 38oF H3 = heat gained by 5 pounds of water (produced by ice after melting) warming from 32oF to 38oF H1 + H3 = H2 H2 = 25 lbs. x 1 x (68oF - 38oF) = 750 BTU's H3 = 5 lbs. x 1 x (38oF - 32oF) = 30 BTU's H1 = H2 - H3 H1 = 750 BTU's - 30 BTU's = 720 BTU's 720 BTU's is the amount of heat required to melt 5 pounds of ice. The latent heat of fusion for 1 pound of ice = 720 BTU's / 5 Therefore, Latent Heat of Fusion of Ice = 144 BTU's per pound. amount of heat gained in fusion of ice = amount of heat lost in crystallization of water
BTU: unit of heat. One BTU is the energy required to raise 1 pound of water by 1 degree Fahrenheit. Specific heat: the ratio of the amount of heat required to increase the temperature of one pound of any substance by one degree to the amount necessary to increase one pound of water. The specific heat of water is 1, by adoption as a standard, and specific heat of another substance (solid, liquid, or gas) is determined experimentally by comparing it to water.
Sh = Specific heat Hq = Heat quantity (BTU) M = Mass (weight in lbs.) t1 - t2 = Temperature change (oF) Large quantities of BTU's are stated as MBTU's: 1,000 BTU = 1 MBTU
Latent Heat of Vaporization After the ice is melted, however, the temperature of the water is raised when more heat is applied. When 180 BTU are added, the water boils. To change a pound of water at 212°F to a pound of steam at 212°F, you must add 970 BTU (latent heat of vaporization). After the water is converted to steam at 212°F, the application of additional heat causes a rise in the temperature of the steam. Example One pound of steam is added to 19 pounds of water. The initial temperature of the water is 40°F, the temperature of the steam is 212°F, and the temperature of the resulting mixture is 97.1°F. Find the latent heat of vaporization of water. Solution H1 = latent heat of vaporization of steam = heat lost by 1 pound of steam changing state from gas at 212°F to water at 212°F H2 = heat gained by 19 pounds of water warming from 40°F to 97.1°F H3 = heat lost by 1 pound of water cooling from 212°F to 40°F H1 + H3 = H2 H2 = 19 lbs. x 1 x (97.1oF - 40oF) = 1084.9 BTU's H3 = 1 lbs. x 1 x (212°F - 97.1°F) = 114.9 BTU's H1 = H2 - H3 H1 = 1084.9 BTU's - 114.9 BTU's = 912.9 BTU's amount of heat lost in condensing = amount of heat gained in vaporization Therefore, Latent Heat of Vaporization of Water = 970 BTU's per pound. |
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