First...What is the difference between HEAT and TEMPERATURE?
Specific Heat Capacity (C or S ) - The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity of the substance. The quantity of heat is frequently measured in units of Joules(J). Another property, the specific heat, is the heat capacity of the substance per gram of the substance. The specific heat of water is 4.18 J/g° C.
Substance | C (J/g oC) |
Air | 1.01 |
Aluminum | 0.902 |
Copper | 0.385 |
Gold | 0.129 |
Iron | 0.450 |
Mercury | 0.140 |
NaCl | 0.864 |
Ice | 2.03 |
Water | 4.18 |
q = m x C x DT
q = m x C x (Tf - Ti)
q = amount of heat energy gained or lost by substance
m = mass of sample
C = heat capacity (J oC-1 g-1 or J K-1 g-1)
Tf = final temperature
Ti = initial temperature
Specific Heat Instructional Videos
Solving for Heat | ||||||
Solving for Mass | ||||||
Solving for Specific Heat | ||||||
Solving for Final Temperature | ||||||
College Level Specific Heat Calculations | ||||||
Solving for the Specific Heat of a Metal that is dropped in water. | ||||||
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Solving for Final Temperature when Ice is added to water | ||||||
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More High School Examples
Highlight Answer Box to reveal answers
Sample Questions | Answers |
1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 46oC. | q = m x C x DT q = 250g x 4.18J/goC x 26oC q = 37,620J or 38kJ |
2. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25o to 60o. | q = m x C x DT C= q/m x DT C = 204.75J /(15g x 35oC ) C= 0.39 J/goC |
3. 216 J of energy is required to raise the temperature of aluminum from 15o to 35oC. Calculate the mass of aluminum. (Specific Heat Capacity of aluminum is 0.90 JoC-1g-1). | q = m x C x DT m= q/C x DT m= 216J/(0.90J/goC x 20oC ) m= 12g |
4. The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (Specific heat capacity of ethanol is 2.44 JoC-1g-1). | q = m x C x DT DT = q/m x C DT = 3240J/(150g x 2.44J/goC) DT = 8.85oC Tfinal= 22oC +8.85oC= 30.9oC |
Even More Practice Questions
Question | Highlight to reveal answers |
1. The temperature of a piece of Metal X with a mass of 95.4g increases from 25.0°C to 48.0°C as the metal absorbs 849 J of heat. What is the specific heat of Metal X? | Answer: 849 J /(95.4g x 23.0°C) 0.387 J/g°C |
2. When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of the olive oil? | Answer: 435 J/(3.4g x 64°C) 2.0 J/g°C |
3. A piece of stainless steel with a mass of 1.55 g absorbs 141 J of heat when its temperature increases by 178°C. What is the specific heat of the stainless steel? | Answer: 141 J/(1.55 g x 178°C) 0.511 J/g°C |
4. How much heat is required to raise the temperature of 250.0 g of mercury by 52°C? | Answer: 250.0 g x 0.140 J/g°C x 52°C 1800 J |
6. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 18°C to 85°C? (Hint: You first need to determine the mass of the water, then calculate q in the requested unit.) | Answer: 1.00kg x 4.18 J/g°C x 67°C 280 kJ |
7. A piece of aluminum with a mass of 100.0 g has a temperature of 20.0°C. It absorbs 1100 J of heat energy. What is the final temperature of the metal? | Answer:1100 J/(100.0 g x 0.902J/g°C)=12.2°C + 20°C= 32.2°C |
8. An unknown metal has a mass of 18.0 g. If the temperature of the metal sample rises from 15.0°C to 40.0°C as the sample absorbs 89.0 J of heat, what is the specific heat of the sample? Now look at your periodic table and choose a metal that is most likely the identity of the sample. | Answer: 89.0 J/(18.0 g x 25.0°C) specific heat = 0.199 J/g°C |
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:
- Determine whether you want to warm up the sample (give it some thermal energy) or cool it down (take some thermal energy away).
- Insert the amount of energy supplied as a positive value. If you want to cool down the sample, insert the subtracted energy as a negative value. For example, say that we want to reduce the sample's thermal energy by 63,000 J. Then Q = -63,000 J.
- Decide the temperature difference between the initial and final state of the sample and type it into the heat capacity calculator. If the sample is cooled down, the difference will be negative, and if warmed up - positive. Let's say we want to cool the sample down by 3 degrees. Then ΔT = -3 K. You can also go to advanced mode to type the initial and final values of temperature manually.
- Determine the mass of the sample. We will assume m = 5 kg.
- Calculate specific heat as c = Q / (mΔT). In our example, it will be equal to c = -63,000 J / (5 kg * -3 K) = 4,200 J/(kg·K). This is the typical heat capacity of water.
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.
- ice: 2,100 J/(kg·K)
- water: 4,200 J/(kg·K)
- water vapor: 2,000 J/(kg·K)
- basalt: 840 J/(kg·K)
- granite: 790 J/(kg·K)
- aluminum: 890 J/(kg·K)
- iron: 450 J/(kg·K)
- copper: 380 J/(kg·K)
- lead: 130 J/(kg·K)
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.
- Find the initial and final temperature as well as the mass of the sample and energy supplied.
- Subtract the final and initial temperature to get the change in temperature (ΔT).
- Multiply the change in temperature with the mass of the sample.
- Divide the heat supplied/energy with the product.
- The formula is C = Q / (ΔT ⨉ m).
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.