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Specific Heat

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 oC. Hence Specific Heat Definition is: the heat required to change temperature of a body per gram mass by 1 oC.


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 = Tf – Ti .Tis final temp of body and Ti is initial temp. units of specific heat are J/gm OC, joule/kg oC,calorie/kg oC etc.

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

Substance

Specific-heat in joule/kg oC

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 oC. 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/kgoC) 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 300oC 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 70OC then calculate specific heat of metal. Given that c(calorimeter)=0.2k calorie/kg OC, 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.