Heat may be defined as energy in transit from a high temperature object to a lower temperature object. Heat is form of energy which produces the change in temperature in any substance. The heat is denoted by **Q**.

The Heat formula is given by

Heat formula is used to find the heat transfer, mass, specific heat or temperature difference if some of the quantities among these are given. Heat is expressed in**Joules (J)**.

The Heat formula is given by

Where,

m is the mass of the body,

C is the specific heat,

$\Delta$ T is the temperature difference.

m is the mass of the body,

C is the specific heat,

$\Delta$ T is the temperature difference.

Heat formula is used to find the heat transfer, mass, specific heat or temperature difference if some of the quantities among these are given. Heat is expressed in

Related Calculators | |

Specific Heat Formula Calculator | Heat Calculator |

Heat Capacity Calculator | Heat Transfer Calculator |

Below are some problems based on Heat which may be helpful for you.

### Solved Examples

**Question 1: **Calculate the heat needed to raise a half kg of iron from 27^{0 }C to 60^{0} C?

** Solution: **

Given: Mass m = 0.5 Kg,

Specific heat of iron C = 0.45 J/g^{0}C,

Temperature difference $\Delta$ T = 70^{0}C - 27^{0}C = 43^{0}C

The heat transfer is given by Q = m $\times$ c $\times$ $\Delta$ T

= 0.5 Kg $\times$ 0.45 $\times$ 10^{3} J/Kg^{0}C $\times$ 43^{0} C

= 9.675 J.

**Question 2: **How much heat energy is lost if water having 2 Kg mass is cooled from 60^{0}C to 20^{0}C?
(Specific heat of water C = 4.2 $\times$ 10^{3} J/Kg^{0}C)

** Solution: **

Mass of water m = 2 Kg,

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

Temperature difference $\Delta$ T = - 40^{0}C

Heat energy is given by Q = m C $\Delta$ T = 2 Kg $\times$ 4.2 $\times$ 10^{3 }J/Kg^{0}C $\times$ (-40^{0} C)

= - 31.6 J.

Given: Mass m = 0.5 Kg,

Specific heat of iron C = 0.45 J/g

Temperature difference $\Delta$ T = 70

The heat transfer is given by Q = m $\times$ c $\times$ $\Delta$ T

= 0.5 Kg $\times$ 0.45 $\times$ 10

= 9.675 J.

Mass of water m = 2 Kg,

Specific heat of water C = 4.2 $\times$ 10

Temperature difference $\Delta$ T = - 40

Heat energy is given by Q = m C $\Delta$ T = 2 Kg $\times$ 4.2 $\times$ 10

= - 31.6 J.