**Specific Heat Capacity**

**Specific heat capacity, c, of a body is defined as the amount of heat (Q) required to raise the temperature (θ) of a unit mass of it by one degree, without going through a change in state.**- When the mass of an object is greater, the object will contain more atoms or molecules than a less massive object made up of the same material. Hence, when the temperature of the objects are raised, the more massive object will require a larger thermal energy than the less massive object. (
*Analogy: The more massive object has more “mouths” to feed*) It is thus more common to consider the heat capacity per unit mass or specific heat capacity of the body.

SI unit of specific heat capacity is joule per kilogram per kelvin (J kg^{-1 }K^{-1}) or joule per kilogram per degree Celsius (J kg^{-1 }°C^{-1})

$Q = mc \Delta \theta$, where

c = specific heat capacity (J kg^{-1 }K^{-1}, J kg^{-1 }°C^{-1})

m = mass of substance (kg)

Q = heat or thermal energy absorbed or released (J)

Δθ = change in temperature (K or °C)

Specific heat capacity of gases is higher than that of liquids and much higher than that of liquids and much higher than that of solids. The substances with higher specific heat capacity cool or warm very slowly compared to substances with lower specific heat capacity.

With gases, the molar heat capacity (the heat capacity of one mole of a gas at constant pressure or constant volume) is generally more useful than the specific heat capacity, which is based on mass.

**Worked Example:**

An electric heating coil supplies 50 W of power to a metal block of mass 0.60 kg and raises the temperature of the block from 20 °C to 45 °C in 90 s. Calculate the specific heat capacity of the metal. What assumption did you make to arrive at your answer?

### Show/Hide Answer

**Assumption:** No heat loss to the surroundings, i.e. all the heat supplied by the heater is absorbed by the metal.

Heat energy supplied by heater, E = P x t = 4500 J

Recall: $E = mc \Delta \theta$

$c = \frac{4500}{0.60 \times 25}$

c = 300 Jkg^{-1}°C^{-1}

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