SfC Home > Physics > Thermal Energy > Show by Ron Kurtus It takes a certain amount of heat energy or thermal energy to turn ice into water and water into steam. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The only exception is when the material reaches its melting point or boiling point. At those two temperatures, the heat energy goes into changing the state or phase of the material. After the state has changed, the temperature will rise again with added thermal energy. The rate temperature changes is the specific heat of the material. The amount of energy required to melt the material is called the latent heat of melting. This all can be illustrated in showing how much heat is required to change ice into water and then change the water into steam. Questions you may have include:
This lesson will answer those questions. Useful tool: Units Conversion Since we are measuring the amount of heat required to make these changes, we need to know the definitions of the various units involved. Heat transferHeat is the total kinetic energy of all the molecules in an object. Although energy is typically measured in joules, a more common unit for heat is the calorie, which is defined as the amount of heat required to change the temperature of 1 gram of water by 1 Celsius degree. There is a conversion factor to relate joules to calories, but we won't worry about that here.
In the English measurement system, they use the BTU (British thermal unit), which is the amount of heat required to change the temperature of 1 pound of water by 1 Fahrenheit degree. The BTU is seen in the United States when referring to the capacity of a furnace. Specific heatMaterials vary in their capacity to store thermal energy. For example, a material like iron will heat up much faster than water or wood. The measurement of how much heat is required to change the temperature of a unit mass of a substance by 1 degree is called its specific heat. There are charts available listing the specific heat of various materials. The chart below shows the specific heat of ice, water and steam in units of calorie per gram-degree Celsius. (Note that all items are listed with the same number of decimal points. That indicates the same accuracy for each. Also, the zero in front of the decimal point assures that the reader will know it is a decimal point and not a fly speck.)
Specific heat of various states of water In other words, it would take twice as many calories to heat some water one degree than it would to heat the same mass of ice one degree. Latent heatWhen any material is heated to the temperature where it changes state, the temperature will remain the same until all the material changes state. That means ice water will remain at 0oC (32oF) until all the ice is melted. The same thing applies when cooling the materials. The reason is that energy must be expended to change the state from solid to liquid or from liquid to gas. Likewise, energy must be withdrawn to change the state when cooling the material. The amount of energy required is call the latent heat of freezing or boiling. The chart below shows the latent heat or energy required to change the states of water.
Latent heat required to change state of water ProblemA good way to understand the concepts is to solve a problem. Suppose we have 50g of ice at -10oC. We want to heat the material until it all turns to steam at 110oC. How much heat is required? With a problem that is complex like this one, it is good to break it down into pieces and solve each part individually. This also helps to explain the logic used in the solution. 1. Heating iceHow much heat would be required to raise 50g of ice to its melting point?The ice temperature must be raised 10 degrees to reach 0oC. Since the specific heat of ice is 0.50 cal/g-oC, that means that 0.50 calories is needed to raise 1g of ice 1oC. Thus, it would take 50 x 0.50 calories to raise 50g up 1oC and 10 x 50 x 0.50 = 250 cal to raise the ice to its melting point. 2. Melting iceHow much heat would be required to melt the 50g of ice?The latent heat for melting ice is 80 cal/g. That means that 1g of ice requires 80 cal of heat to melt. Thus, 50g requires 50 x 80 = 4000 cal to melt. 3. Heating waterHow much heat is required to heat 50g of water from 0oC to its boiling point of 100oC?Since the specific heat of water is 1.00 cal/g-oC, that means that 1.00 calorie is needed to raise 1g of water 1oC. Thus, it would take 50 x 1.00 calories to raise 50g up 1oC and 100 x 50 x 1.00 = 5000 cal to raise the water to its boiling point. 4. Boiling waterHow much heat would be required to boil the 50g of water?The latent heat for boiling water is 540 cal/g. That means that 1g of water requires 540 cal of heat to boil. Thus, 50g requires 50 x 540 = 27000 cal to boil. 5. Heating steamHow much heat is required to heat 50g of steam from 100oC to 110oC?Since the specific heat of steam is 0.48 cal/g-oC, that means that 0.48 calories are needed to raise 1g up 1oC. Thus, it would take 50 x 0.48 calories to raise 50g of steam 1oC and 10 x 50 x 0.48 = 240 cal to raise the temperature of the steam to 110oC. 6. TotalThe total heat required to change 50g of ice at -10oC to steam at 110oC is: 250 + 4000 + 5000 + 27000 + 240 = 36490 cal. SummaryHeating materials like ice, water and steam increases their temperature. The specific heat of the material determines the calories required to heat one unit of mass one degree. Changing the state of the material requires extra heat energy that is not used to change the temperature. The amount of heat required to change the state of the material is called its latent heat. The complex problem of determining how much heat is required to change ice into water and then change the water into steam should be broken into parts and solved individually. Heat up your thinking Resources and referencesRon Kurtus' Credentials WebsitesSpecific Heat Equations - Wolfram Science World Physics Resources Books(Notice: The School for Champions may earn commissions from book purchases) Top-rated books on Thermal Energy Top-rated books on Physics of Temperature Share this pageClick on a button to bookmark or share this page through Twitter, Facebook, email, or other services: Students and researchersThe Web address of this page is: Please include it as a link on your website or as a reference in your report, document, or thesis. Copyright © Restrictions Where are you now?School for Champions
Heat of fusion is the amount of heat energy required to change the state of matter of a substance from a solid to a liquid. It's also known as enthalpy of fusion. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice.
What is the heat in Joules required to melt 25 grams of ice? What is the heat in calories? Useful information: Heat of fusion of water = 334 J/g = 80 cal/g In the problem, the heat of fusion is given. This isn't a number you're expected to know off the top of your head. There are chemistry tables that state common heat of fusion values.
To solve this problem, you'll need the formula that relates heat energy to mass and heat of fusion: q = heat energy m = massΔHf = heat of fusion Temperature is not anywhere in the equation because it doesn't change when matter changes state. The equation is straightforward, so the key is to make sure you're using the right units for the answer. To get heat in Joules:q = (25 g)x(334 J/g)q = 8350 JIt's just as easy to express the heat in terms of calories: q = m·ΔHf q = (25 g)x(80 cal/g)q = 2000 calAnswer: The amount of heat required to melt 25 grams of ice is 8,350 Joules or 2,000 calories. Note: Heat of fusion should be a positive value. (The exception is helium.) If you get a negative number, check your math. |
What is the amount of heat needed to completely melt 30 grams of ice?
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