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Physics Formulas

Heat Capacity Formula

Heat capacity is the core concept of thermodynamics. Furthermore, in this topic, you will learn about heat capacity, heat capacity formula, formula’s derivation, and solved example. Moreover, after completing this topic you will easily be able to understand heat capacity.

heat capacity formula

Heat capacity

Have you ever gone to a beach and become aware of that the sand is really hot in the day and cold at the night, but the temperature of the ocean does not seem to change at all? Also, both are subjected to the same amount of sunlight (energy) from the sun but the different properties of water and sand cause them to react differently.

Furthermore, this happens because the water has much higher heat capacity than the sand. Besides, it simply means that it takes much more energy to raise the temperature of water than it does for sand.

Understanding Heat

For understanding heat capacity firstly we need to understand heat. Heat is a kind of energy that passes from objects of higher temperature to object of lower temperature. For example, if we touch a hot mug of coffee then we will feel hot because the mug transfers its energy (heat) into our body.

On the other hand, if we touch a glass of cold water then the heat energy transfer from our body to the glass making it feels cold.
Furthermore, the SI unit for heat is Joules (J). In addition, it is noted as Q in the equation.

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Defining Heat Capacity

It is the amount of energy that we require to raise the temperature of a specific substance by one degree Celsius. Moreover, it can also be viewed as the ratio of the amount of energy transferred to an object and the resulting temperature rise.

Furthermore, the heat capacity unit is Joules per degree Celsius. Since it depends on the mass of the object. Moreover, it is often given in per 100 grams to allow for a comparison between the object of equal mass.

Heat Capacity Formula

Heat capacity is the thermal mass of the object and is defined as the energy in Joules required raising the temperature of the given object by one degree Celsius. Furthermore, this specific heat of the object (defined chemical/physical property) multiplied by its mass and the change in temperature.

Heat capacity = mass × specific heat × change in temperature
Q = mc \(\Delta\)T

Derivation of the Heat Capacity Formula

Q = refers to the specific heat in Joules (J)
m = refers to the mass of the object in grams (g)
c = refers to the specific heat of the object in joules per gram degree Celsius (J/\((g^{-\circ}C)\))
\(\Delta\)T = refers to the change in temperature in degree Celsius (\(^{\circ}C\))

Solved Example

Example 1

A piece of iron 125 g has a specific heat = 0.45 \(J/g^{\circ}C\). Also, it is heated from \(100^{\circ}C\) to \(450^{\circ}C\). So, calculate how much heat energy is required?

Solution:

mass (m) = 125 g
specific heat of iron, (c) = 0.45\(J/g^{\circ}C\)
change in temperature (\(\Delta\)T) = 450 – 100 = \(350^{\circ}C\)

Calculation

Q = mc \(\Delta\)T
Q = (125 g) (0.45\(J/g^{\circ}C\)) (\(350^{\circ}C\))
Q = 19687.5 J

So, the heat capacity of 125 g of iron is 19687.5 J.

Example 2

Suppose that 15,245 J of heat is applied to a copper ball of a mass 45 g. Also, the specific heat of copper is = 0.39\(J/g^{\circ}C\). So, find how much will the temperature change?

Solution:

mass (m) = 45 g
specific heat of copper (c) = 0.39\(J/g^{\circ}C\)
Q = 15245 J

Calculation

Q = mc \(\Delta\)T \(\rightarrow\) \(\frac{Q}{mc}\) = \(\Delta\)T

\(\Delta\)T = \(\frac{15245 J}{(45 g) (0.39J/g^{\circ}C)}\)

\(\Delta\)T = \(868.66^{\circ}C\)

So, the temperature change is about \(868.66^{\circ}C\).

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