What is the term for the amount of energy needed to raise the temperature of 1kg of water by 1 C?

Hello, and welcome to this Mometrix video on specific heat capacity—a constant that relates heat transfer to changes in temperature.

Temperature is directly related to the average translational kinetic energy of the atoms or molecules in a system. Basically, the faster and heavier the particles are, the higher the temperature. The units for temperature are degrees Celsius and Kelvin (remember, 0 degrees Celsius = 273 Kelvin, but 1 degree Celsius has the same magnitude as 1 Kelvin).

Conversely, heat is measured in joules and is the energy transferred between systems at different temperatures that are in contact. Because heat is the transfer of energy, it is known as a process quantity.

When heat is absorbed or released by a system, the temperature changes. How much the temperature changes depends on the substance, and specifically, the specific heat capacity of that substance.

Let’s look at an example.

Let’s say we have 47.8 grams of water at 35ºC and we put it on the stove. We turn on the stove and transfer 1,000 joules of heat to our water and the temperature rises to 40ºC. In other words, it took 1,000 joules of heat to raise 47.8 grams of water by 5ºC.

That’s kind of a mouthful and seems oddly specific in terms of quantities. This is where specific heat capacity, notated as \(c\), comes into play. It’s a standard quantity and is the amount of heat required to raise 1 gram of a substance by 1 degree Celsius.

With some simple division, we can derive the specific heat capacity of water from our hypothetical cup of water.

\(\text{Specific heat capacity (}c\text{)} =\) \(qmass \times ΔT= \frac{1000\text{ joules}}{47.8\text{ g} \times 5\text{ K}}\)\(=4.184\text{ J/gK}\)

From this, we know now that it takes 4.184 joules to raise the temperature of 1 gram of water by 1 degree Celsius—that’s the specific heat capacity of water.

This is really helpful to scientists because once determined, the specific heat capacity can be used to calculate the heat absorbed or released by a system simply by measuring the temperature change and mass. Let’s try that out.

Let’s look at a new cup of water, let’s say a mug of 350 grams that’s boiling. The water starts at 100ºC and cools down to 90ºC. We want to know how much heat was released from the water to the surrounding environment. Since we know the specific heat capacity of water is 4.184 joules per gram Kelvin, we simply need to rearrange our previous equation to solve for \(q\) (the heat released).

\(q=c \times \text{mass} \times ∆T\)\(=4.184 \text{ J/gK} \times 350 \text{g} \times -10 \text{ K}\)\(=-14,644 \text{ J}\)

Since we knew the specific heat capacity of water, calculating the heat released from the system was easy!

Note here that the negative sign simply tells us that the system (the mug of water) released heat to the surrounding system rather than absorbed it.

Now that we’ve defined specific heat capacity and demonstrated how it can be used to calculate the heat transferred from or to a system, let’s look at why the specific heat capacity changes between substances.

For example, the specific heat capacity of ethanol is 2.18 joules per gram Kelvin, almost half of water. If we have one gram of water and one gram of ethanol both at 0ºC, it would take 4.18 joules of heat to raise the temperature of water to 1ºC, and only 2.18 joules for ethanol. The liquids reach the same temperature but require different amounts of heat. Why?

Remember, to increase the temperature, we need to increase the average translational kinetic energy of the molecules (make the molecules move faster). But the internal energy of a substance is more than just the translational kinetic energy, it also includes potential energy from intermolecular interactions. When heat is transferred to a system, it is distributed amongst the kinetic and potential energies.

So, if a system has more potential energy, a smaller proportion of the transferred heat is distributed to the kinetic energy, yielding a smaller increase in temperature. To better understand this concept, let’s look at water and ethanol again.

In water, there is a complex network of hydrogen bonds between the molecules. Those interactions are part of the potential energy and need to be overcome, or broken, to increase the average translational kinetic energy. So, when we heat water, some of that energy is used to break up the hydrogen bonding network instead of increasing the kinetic energy, resulting in a large specific heat capacity. Conversely, in ethanol, there are fewer hydrogen bonds per molecule, or less potential energy, and therefore a larger proportion of the heat transferred is used to increase the average kinetic energy, which results in a smaller specific heat capacity.

Review

Okay, let’s wrap up with a review. First, we reviewed the scientific definitions of temperature and heat and related them using specific heat capacity. Using water as an example, we showed how once we know the specific heat capacity, it is quite easy to determine the heat transferred from or to a system. And finally, we considered from a microscopic view why substances have different specific heat capacities.

Thanks for watching and happy studying!

Specific heat capacity refers to the amount of energy or heat required to increase the temperature of 1 gram of a substance by one degree Celsius.

A key characteristic of water is its high specific heat capacity. Water has to absorb 4,184 Joules of heat, or 1 calorie, to increase in temperature of 1 kilogram of water by 1 degree Celsius.

The units often expressed for specific heat capacity are J/(g×C) or J/(kg×C).

Heat capacity, also known as thermal mass, refers to the amount of heat energy require to raise the temperature of an object, and is measure in Joules per Kelvin or Joules per degree Celsius. Specific heat capacity, also known as specific heat, is the heat capacity per unit mass.

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:

1. Determine whether you want to warm up the sample (give it some thermal energy) or cool it down (take some thermal energy away).
2. 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.
3. 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.
4. Determine the mass of the sample. We will assume m = 5 kg.
5. 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:

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.

1. Find the initial and final temperature as well as the mass of the sample and energy supplied.
2. Subtract the final and initial temperature to get the change in temperature (ΔT).
3. Multiply the change in temperature with the mass of the sample.
4. Divide the heat supplied/energy with the product.
5. 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.