 # Specific Heat Formula

Specific Heat is used when we are referring to something particular. Heat capacity is the ratio of the quantity of heat required to alter the temperature by one degree Celsius. But when we consider a certain amount of mass we make use of the word Specific Heat or Specific Heat Capacity.

Specific Heat is the quantity of heat essential to raise the temperature of a gram of any substance by 1 degree Celsius.

Specific Heat formula is articulated as

$C\;=\;\frac{\Delta Q}{m\;\Delta T}$

Where,
Δ Q is the heat gained or lost,
Δ T is the temperature difference.

The temperature difference is given by Δ T = (Tf – Ti), where the final temperature is Tf  and the initial temperature is Ti .

Specific Heat formula is made use of to find the specific heat of any given material, its mass, heat gained or temperature difference if some of the variables are given. It is articulated in Joule/Kg Kelvin (J/Kg K).

Specific Heat Solved Examples

Underneath are some questions based on specific heat which will be useful for you.

Problem 1: Calculate the heat required to raise 0.6 Kg of sand from 30oC to 90oC? (Specific Heat of sand = 830 J/KgoC)

Known:

Mass of sand m = 0.6 Kg,
Δ T (Temperature difference)  = 90oC – 30oC = 60oC
C (Specific Heat of sand) = 830 J/KgoC

The specific heat is given by, $C\;=\;\frac{\Delta Q}{m\;\Delta T}$

Henceforth, Heat required is given by Q = C m Δ T
= 830 J/KgoC × 0.6 Kg × 60oC
= 29880 J.

Problem 2: Compute the temperature difference if 50 Kg of water absorbs 500 K J of heat?
$\Delta T\;=\;\frac{Q}{C\;m}\;=\;\frac{500\times 10^{3}}{4.3\times 10^{3}\times 50}$