When we give heat to a body its temperature raises and when heat is taken from the body its temperature decreases. Heat capacity is the amount of heat which is needed to change the temperature of a body by certain amount. While specific heat(heat) is the heat required to change temperature of one gram of a body by 1 ^{o}C. Hence Specific Heat Definition is: the heat required to change temperature of a body per gram mass by 1 ^{o}C.

**Specific Heat Equation is given as:**

Q = m c ∆t

Here Q is the amount of heat required to for a body of mass m to change its temperature by ∆t. here ∆t is change in temp which is given by the difference of final temp and initial temp. ∆t = T_{f} – T_{i} .T_{f }is final temp of body and T_{i} is initial temp. units of specific heat are J/gm ^{O}C, joule/kg ^{o}C,calorie/kg ^{o}C etc.

Let us see specific heats of some of the materials in Specific Heat Table given below:

Substance | Specific-heat in joule/kg |

Silver | 230 |

Copper | 390 |

Aluminum | 900 |

Lead | 130 |

Air | 718 |

wood | 2100 |

Steam | 2000 |

Water | 4186 |

Steel | 490 |

Glass | 670 |

Platinum | 130 |

Specific Heat of Silver as given above is 230 joule/kg ^{o}C. This means it takes 230 joules of heat to change temperature of 1 kg of silver by 1 degree Celsius. Temp change can be positive or negative according to the release or absorption of heat. A similar conclusion can be taken for Specific Heat of Steel (490J/kg^{o}C) and other metals.

Is this topic **Heat Capacity Formula** have hard for you? Watch out for my coming posts.

Calorimetry is the technique of measurement of heat involved in a chemical or physical change. Let us deal with some Calorimetry Problems now:

Q.1) A 1 kg piece of some metal at 300^{o}C is placed in 0.1 kg of water at freezing point of water in a 0.05kg calorimeter. If final temp of the system is 70^{O}C then calculate specific heat of metal. Given that c(calorimeter)=0.2k calorie/kg ^{O}C, c(water)= 1k cal/kg°C

Solution) for any physical change in temp:

Heat lost by the system = heat gainedby the system

Here metal will lose heat and water and calorimeter gains heat. so,

heat lost by metal = heat gained by water + heat gained by calorimeter……….(1)

We know that Q="mc" ∆t

heat lost by metal = -1(c)(70-300) (-sign for heat loss)

heat gain by water="0.1(1)(70-0)

heat gain by calorimeter="0.05(0.2)(70-0)

Using equation (1) and putting above values:

-1(c)(70-300)=(0.1)(1)(70-0)+(0.05)(0.2)(70-0)

On solving we get, specific heat of metal =c=3.3 x 10^{-2} kcal/kg°C.