Specific Heat Capacity or Specific Heat [c]:-It has been observed that different bodies, of same mass required different amounts of heat to raise their temperatures to same level. It is an experimental fact that,

(a) greater the mass ‘m’ of body, greater heat is required to raise its temperature by same amount, so quantity of heat Q is, Q ∝ m

(b) greater heat is required to raise the temperature higher, Q ∝ ∆T

Here ∆T is the rise in temperature.

Combining the two factors together,

Q ∝ m ∆T

Or, Q = cm ∆T

Here c is called the “specific heat” or “specific heat capacity” of the body. It depends only on the nature of material.

c = Q / m ∆T

If m = 1, ∆T = 1°C, c = Q

Specific heat capacity of a material is defined as the amount of heat required to raise the temperature of a unit mass of material through 1°C.

Actually ‘c’ is the mean specific heat capacity over a temperature range ∆T, since we know that the quantity of heat required to raise the temperature of material through a small interval varies with the location of the interval in the temperature scale. If ‘∆Q’ is the small amount of heat required to increase the temperature by a small amount of temperature ‘∆T’, the true specific heat capacity is defined as,

c = 1/m (dQ/dT)

To calculate ‘Q’ we shall have to perform integration.

Q = ∫ dQ = m ∫ c dT

‘c’ being a function of ‘T’.

Dimension of c:-c = (energy) / (mass) (temperature)

= [M

^{0}L^{2}T^{-2}K^{-1}]

Units of specific heat:-kcal kg

^{-1}K^{-1}or J kg^{-1}K^{-1}

Molar specific heat capacity:-Molar specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of one gram molecule of the substance through one degree centigrade. It is denoted by C.

One mole of substance contains M gram of substance where M is the molecular weight of the substance.

So, C = Mc

If n is the number of moles of substance, then,

n = m/M

So, m = nM

Substituting for m in equation c = 1/m (dQ/dT), we get,

c = 1/nM (dQ/dT)

Or, Mc = 1/n (dQ/dT)

Thus, C = Mc

= 1/n (dQ/dT)

Specific heat of water is taken to be 1. This is because of the reason that we defined unit of heat (calorie) by making use of water.

Heat capacity or Thermal capacity:-## It is defined as the amount of heat required to raise the temperature of body through 1ºC.

Q=mcΔTIf Δ

T= 1ºC,Q= heat capacity =mcThus, heat capacity of a body is equal to the product of mass and its specific heat capacity.

Unit:-kcalK^{-1}orJK^{-1}

Water Equivalent:-Consider a body of mass ‘m’ g and water of mass ‘w’ g. Supply same quantity ‘Q’ to both of them. If both of them register same rise of temperature (θ), ‘w’ is said to be the water equivalent of the body.

Water equivalent of a body is defined as the mass of water which gets heated through certain range of temperature by the amount of heat required to raise the temperature of body through same range of temperature.

For the body, Q = mcθ

For water, Q = w×1×θ

= wθ

So, wθ = mcθ

Or, w = mc

Water equivalent of a body is equal to the product of its mass and its specific heat.

Dimension:-[M^{1}L^{0}T^{0}]

Units:-kg## Latent Heat:-

When the state of matter changes, the heat absorbed or evolved is given by:

Q=mL. HereLis called the latent heat. The magnitude of latent heat depends upon the mass of the substance.

Dimension of latent heat:-[M^{1}L^{2}T^{-2}]

Units of latent heat:-kcal or Joule

Specific latent heat:-Corresponding to the two stages of conversion (fusion and vaporization), we have two categories of specific latent heat.

(a) Specific latent heat of fusion (

L):- Specific latent heat of fusion of a substance is defined as the amount of heat required to convert 1 gram of substance from solid to liquid state, at the melting point, without any change of temperature._{f}(b) Specific latent heat of vaporization (

L):- Specific latent heat of vaporization of a substance is defined as the amount of heat required to convert 1 gram of liquid into its vapors at its boiling point without any rise of temperature._{v}

Dimensional formula:-M^{0}L^{2}T^{-2}

Unit:-kg cal kg^{-1 }orJ kg^{-1}

Problem 1:-In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloud days, 5.22 GJ are needed to maintain the inside of the house at 22.0°C, Assuming that the water in the barrels is at 50.0°C, what volume of water is required?

Concept:-Heat

Qthat must be given to a body of massm, whose material has a specific heatc, to increase its temperature from initial temperatureT_{i}to final temperatureT_{f}is,

Q=mc(T_{f}-T_{i})So,

m=Q/ c(T_{f}-T_{i})Density

ρis equal to massmper unit volumeV.So,

ρ=m/VSo volume

Vwill be,

V=m/ρ

Solution:-To find the volume water, first we have to find out the mass of water which is required to transfer 5.22 GJ amount of heat energy.

To find the mass

mof water, substitute 5.22 GJ forQ, 4190 J/kg. K for specific heat capacitycof water, 50.0^{̊}C forT_{f}and 22.0 ̊ C forT_{i}in the equationm=Q/ c(T_{f}-T_{i}),

m=Q/ c(T_{f}-T_{i})= 5.22 GJ/(4190 J/kg. K) (50.0

^{̊}C-22.0 ̊ C)= (5.22 GJ) (10

^{9}J/1 GJ)/(4190 J/kg. K) ((50.0+273) K –(22.0+273) K)= (5.22 ×10

^{9}J)/(4190 J/kg. K) (28 K)= 4.45×10

^{4}kgTo obtain the volume

Vof water, substitute 4.45×10^{4}kg for massmand 998 kg/m^{3}for densityρof water in the equationV=m/ρ,

V=m/ρ= (4.45×10

^{4}kg) / (998 kg/m^{3})= 44.5 m

^{3}From the above observation we conclude that, the volume

Vof water will be 44.5 m^{3}.

Problem 2:-How much water remains unfrozen after 50.4 kJ heat have been extracted from 258 g of liquid water initially at 0°C?

Concept:-The amount of heat per unit mass that must be transferred to produce a phase change is called the latent

Lfor the process. The total heat transferred in a phase change is then

Q=Lm.Here

mis the mass of the sample that changes phase. The heat transferred during melting or freezing is called the heat of fusion.So from the above equation

Q=Lm, mass of the substance would be,

m=Q/L

Solution:-To obtain the amount of water (

m) which freezes, substitute 50.4 kJ for the heatQand 333 ×10^{3}J/kg for latent heat of fusion of water in the equationm=Q/L,

m=Q/L= 50.4 kJ/ (333 ×10

^{3}J/kg)= (50.4 kJ×10

^{3}J/1 kJ) / (333 ×10^{3}J/kg)= 0.151 kg

So the amount of water (

mʹ) which remains unfrozen will be,

mʹ =258 g – 0.151 kg= (258 g×10

^{-3}kg/1 g) –(0.151 kg)= 0.258 kg-0.151 kg

= 0.107 kg

From the above observation we conclude that, the amount of water which remains unfrozen would be 0.107 kg.

Problem 3:-An aluminum electric kettle of mass 0.560 kg contains a 2.40-kW heating element. It is filled with 0.640 L of water at 12.0°C. How long will it take (a) for boiling to begin and (b) for the kettle to boil dry? (Assume that the temperature of the kettle does not exceed 100°C at any time)

Concept:-The amount of heat per unit mass that must be transferred to produce a phase change is called the latent

Lfor the process. The total heat transferred in a phase change is then

Q=Lm,Here

mis the mass of the sample that changes phase. The heat transferred during melting or freezing is called the heat of fusion.The heat which is given to a body of mass

m, whose material has a specific heatc, to increase its temperature from initial temperatureT_{i}to final temperatureT_{f}will be,

Q=mc(T_{f }-T_{i})=

mcΔTMass

mis equal to densityρof object times volumeVof the object.

m=ρVTime

tis equal to the heat energyQdivided by powerP.

t=Q/P

Solution:-(a) The time

ttaken by the electric kettle for boiling to begin will be,

t=Q/P= [

m_{a}c_{a}+ρ_{w}V_{w}c_{w}]ΔT/P= [

m_{a}c_{a}+ρ_{w}V_{w}c_{w}] (T_{f }-T_{i})/PHere, mass of aluminum electric kettle is

m_{a}, specific heat of aluminum isc_{a}, density of water isρ_{w}, volume of water isV_{w}, specific heat of aluminum isc_{w}, final temperature isT_{f}, initial temperature isT_{i}and power isP.To obtain the time

ttaken by the electric kettle for boiling to begin, substitute 0.56 kg form_{a}, 900 J/kg. K forc_{a}, 998 kg/m^{3}forρ_{w}, 0.640 L forV_{w}, 4190 J/kg.K forc_{w}, 100̊C forT_{f}and 12^{̊}C forT_{i}and 2.40 kW forPin the equationt=[m_{a}c_{a}+ρ_{w}V_{w}c_{w}] (T_{f }-T_{i})/P,

t=[m_{a}c_{a}+ρ_{w}V_{w}c_{w}] (T_{f }-T_{i})/P= [(0.56 kg) (900 J/kg. K) + (998 kg/m

^{3}) (0.640 L) (4190 J/kg.K)] (100̊C -12^{̊}C)/2.40 kW= [(0.56 kg) (900 J/kg. K) + (998 kg/m

^{3}) (0.640 L×10^{-3}m^{3}/1 L) (4190 J/kg.K)] ((100+273) K –(12+273) K)/(2.40 kW×10^{3}W/1 kW)= [(0.56 kg) (900 J/kg. K) + (998 kg/m

^{3}) (0.640×10^{-3}m^{3}) (4190 J/kg.K)] (373 K –285 K)/(2400 W)= 117 s

From the above observation we conclude that, the time t taken by the electric kettle for boiling to begin would be 117 s.

(b) The time

tfor the kettle to boil dry will be,

t=Q/P=

ρ_{w}V_{w}L_{w}/PHere density of water is

ρ_{w}, volume of water isV_{w}, latent heat of water isL_{w}and power isP.To obtain the time

tfor the kettle to boil dry, substitute 998 kg/m^{3}forρ_{w}, 0.640 L forV_{w}, 2256×10^{3}J/kg forL_{w}and 2.40 kW forPin the equationt=ρ_{w}V_{w}L_{w}/P,

t=ρ_{w}V_{w}L_{w}/P= (998 kg/m

^{3}) (0.640 L) (2256×10^{3}J/kg)/( 2.40 kW)= (998 kg/m

^{3}) (0.640 L×10^{-3}m^{3}/1 L) (2256×10^{3}J/kg)/( 2.40 kW×10^{3}W/1 kW)= (998 kg/m

^{3}) (0.640 ×10^{-3}m^{3}) (2256×10^{3}J/kg)/( 2400 W)= 600 s

From the above observation we conclude that, the time t for the kettle to boil dry would be 600 s.

Related Resources:You might like to refer some of the related resources listed below:Click here for the Detailed Syllabus of IIT JEE Physics.

Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.

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