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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 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.
Materials 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.
When 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
A 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 ice
How 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 ice
How 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 water
How 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 water
How 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 steam
How 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.
The total heat required to change 50g of ice at -10oC to steam at 110oC is:
250 + 4000 + 5000 + 27000 + 240 = 36490 cal.
Heating 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 references
Ron Kurtus' Credentials
Specific Heat Equations - Wolfram Science World
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Top-rated books on Thermal Energy
Top-rated books on Physics of Temperature
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This specific heat calculator is a tool that determines the heat capacity of a heated or a cooled sample. Specific heat is the amount of thermal energy you need to supply to a sample weighing 1 kg to increase its temperature by 1 K. Read on to learn how to apply the heat capacity formula correctly to obtain a valid result.
💡 This calculator works in various ways, so you can also use it to, for example, calculate the heat needed to cause a temperature change (if you know the specific heat). If you have to achieve the temperature change in a determined time, use our watts to heat calculator to know the power required. To find specific heat from a complex experiment, calorimetry calculator might make the calculations much faster.
Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you:
If you have problems with the units, feel free to use our temperature conversion or weight conversion calculators.
The formula for specific heat looks like this:c = Q / (mΔT)
Q is the amount of supplied or subtracted heat (in joules), m is the mass of the sample, and ΔT is the difference between the initial and final temperatures. Heat capacity is measured in J/(kg·K).
You don't need to use the heat capacity calculator for most common substances. The values of specific heat for some of the most popular ones are listed below.
Having this information, you can also calculate how much energy you need to supply to a sample to increase or decrease its temperature. For instance, you can check how much heat you need to bring a pot of water to the boil to cook some pasta.
Wondering what the result actually means? Try our potential energy calculator to check how high you would raise the sample with this amount of energy. Or check how fast could the sample move with this kinetic energy calculator.
The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m).
The formula for specific heat capacity, C, of a substance with mass m, is C = Q /(m ⨉ ΔT). Where Q is the energy added and ΔT is the change in temperature. The specific heat capacity during different processes, such as constant volume, Cv and constant pressure, Cp, are related to each other by the specific heat ratio, ɣ= Cp/Cv, or the gas constant R = Cp - Cv.
Specific heat capacity is measured in J/kg K or J/kg C, as it is the heat or energy required during a constant volume process to change the temperature of a substance of unit mass by 1 °C or 1 °K.
The specific heat of water is 4179 J/kg K, the amount of heat required to raise the temperature of 1 g of water by 1 Kelvin.
Specific heat is measured in BTU / lb °F in imperial units and in J/kg K in SI units.
The specific heat of copper is 385 J/kg K. You can use this value to estimate the energy required to heat a 100 g of copper by 5 °C, i.e., Q = m x Cp x ΔT = 0.1 * 385 * 5 = 192.5 J.
The specific heat of aluminum is 897 J/kg K. This value is almost 2.3 times of the specific heat of copper. You can use this value to estimate the energy required to heat a 500 g of aluminum by 5 °C, i.e., Q = m x Cp x ΔT = 0.5 * 897* 5 = 2242.5 J.