# Heat Formula

Heat is the transfer of kinetic energy from one medium or object to another or via energy source to a medium or object. This energy transfer can occur in three ways namely radiation, conduction and convection. Heat is a form of energy that produces change in temperature of any substance.

Heat by conduction takes place when two objects are in direct contact and temperature of one is higher than the other. The temperature tends to equalize that’s the reason the heat conduction consists of the transfer of kinetic energy from warmer medium to a cooler one. The heat is denoted by Q. Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass

The Specific Heat formula is given as

$c=\frac{Q}{m\times \Delta T}$

Where,

m = mass of the body,

C = specific heat,

Δ T is the temperature difference.

Heat formula can be applied to find the heat transfer, mass, specific heat or temperature difference.

Heat is expressed in Joules (J).

## Solved Examples

Determine the heat needed to raise a half kg of iron from 250 C to 600 C?

Solution:

Given parameters are

Mass m = 0.5 Kg,

Specific heat of iron c= $0.43\times 10^{3}J/Kg^{0}C$

, Temperature difference $\Lambda T=70^{0}-25^{0}=45^{0}C$

the heat formula is given by

Q = m × c × Δ T

$Q=0.5\times 0.43\times 10^{3}\times 45$

= 10.125 x $10^{3}$ J

Example 2

Determine how much heat energy is lost if water of 5 Kg mass is cooled from $60^{0}$

C to $20^{0}$C. Given Specific heat of water C = $4.2\times 10^{3}J/Kg^{0}C$

Solution:

Given parameters are,

Mass of water is 5 Kg,

Specific heat of water C is $4.2\times 10^{3}J/Kg^{0}C$

Temperature difference Δ T is =$40^{0}$

Heat energy formula is given by

Q = m x C x Δ T

= 5 × 4.2 ×$10^{3}$× 40

= 84.0 × $10^{3}$J