Click to Chat
0120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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 = mcΔT If ΔT = 1ºC, Q = heat capacity = mc Thus, heat capacity of a body is equal to the product of mass and its specific heat capacity. Unit:- kcal K^{-1} or JK^{-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. Here L is 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_{f}):- 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. (b) Specific latent heat of vaporization (L_{v}):- 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. Dimensional formula:- M^{0}L^{2}T^{-2} Unit:- kg cal kg^{-1 } or J 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 Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} is, Q = mc (T_{f} - T_{i}) So, m = Q/ c (T_{f} - T_{i}) Density ρ is equal to mass m per unit volume V. So, ρ = m/V So volume V will 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 m of water, substitute 5.22 GJ for Q, 4190 J/kg. K for specific heat capacity c of water, 50.0 ^{°} C for T_{f} and 22.0 ° C for T_{i} in the equation m = 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} kg To obtain the volume V of water, substitute 4.45×10^{4} kg for mass m and 998 kg/m^{3} for density ρ of water in the equation V = 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 V of 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 L for the process. The total heat transferred in a phase change is then Q = Lm. Here m is 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 heat Q and 333 ×10^{3} J/kg for latent heat of fusion of water in the equation m = 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 (m') which remains unfrozen will be, m' =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 L for the process. The total heat transferred in a phase change is then Q = Lm, Here m is 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 heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} will be, Q= mc (T_{f }- T_{i}) = mcΔT Mass m is equal to density ρ of object times volume V of the object. m = ρV Time t is equal to the heat energy Q divided by power P. t = Q/P Solution:- (a) The time t taken 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})/P Here, mass of aluminum electric kettle is m_{a}, specific heat of aluminum is c_{a}, density of water is ρ_{w}, volume of water is V_{w}, specific heat of aluminum is c_{w}, final temperature is T_{f} , initial temperature is T_{i} and power is P. To obtain the time t taken by the electric kettle for boiling to begin, substitute 0.56 kg for m_{a}, 900 J/kg. K for c_{a}, 998 kg/m^{3} for ρ_{w}, 0.640 L for V_{w}, 4190 J/kg.K for c_{w}, 100°C for T_{f} and 12^{°} C for T_{i} and 2.40 kW for P in the equation t =[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)] (100°C -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 t for the kettle to boil dry will be, t = Q/P = ρ_{w}V_{w}L_{w}/P Here density of water is ρ_{w}, volume of water is V_{w}, latent heat of water is L_{w} and power is P. To obtain the time t for the kettle to boil dry, substitute 998 kg/m^{3} for ρ_{w}, 0.640 L for V_{w}, 2256×10^{3} J/kg for L_{w} and 2.40 kW for P in the equation t = 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. To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
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 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.
Q = mcΔT
If ΔT = 1ºC, Q = heat capacity = mc
Thus, heat capacity of a body is equal to the product of mass and its specific heat capacity.
Unit:- kcal K^{-1} or JK^{-1}
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
When the state of matter changes, the heat absorbed or evolved is given by: Q = mL. Here L is 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
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_{f}):- 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.
(b) Specific latent heat of vaporization (L_{v}):- 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.
Dimensional formula:- M^{0}L^{2}T^{-2}
Unit:- kg cal kg^{-1 } or J kg^{-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?
Heat Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} is,
Q = mc (T_{f} - T_{i})
So, m = Q/ c (T_{f} - T_{i})
Density ρ is equal to mass m per unit volume V.
So, ρ = m/V
So volume V will be,
V = m/ρ
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 m of water, substitute 5.22 GJ for Q, 4190 J/kg. K for specific heat capacity c of water, 50.0 ^{°} C for T_{f} and 22.0 ° C for T_{i} in the equation m = 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} kg
To obtain the volume V of water, substitute 4.45×10^{4} kg for mass m and 998 kg/m^{3} for density ρ of water in the equation 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 V of water will be 44.5 m^{3}.
How much water remains unfrozen after 50.4 kJ heat have been extracted from 258 g of liquid water initially at 0°C?
The amount of heat per unit mass that must be transferred to produce a phase change is called the latent L for the process. The total heat transferred in a phase change is then
Q = Lm.
Here m is 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
To obtain the amount of water (m) which freezes, substitute 50.4 kJ for the heat Q and 333 ×10^{3} J/kg for latent heat of fusion of water in the equation 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 (m') which remains unfrozen will be,
m' =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.
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)
Q = Lm,
The heat which is given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} will be,
Q= mc (T_{f }- T_{i})
= mcΔT
Mass m is equal to density ρ of object times volume V of the object.
m = ρV
Time t is equal to the heat energy Q divided by power P.
t = Q/P
(a) The time t taken by the electric kettle for boiling to begin will be,
= [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})/P
Here, mass of aluminum electric kettle is m_{a}, specific heat of aluminum is c_{a}, density of water is ρ_{w}, volume of water is V_{w}, specific heat of aluminum is c_{w}, final temperature is T_{f} , initial temperature is T_{i} and power is P.
To obtain the time t taken by the electric kettle for boiling to begin, substitute 0.56 kg for m_{a}, 900 J/kg. K for c_{a}, 998 kg/m^{3} for ρ_{w}, 0.640 L for V_{w}, 4190 J/kg.K for c_{w}, 100°C for T_{f} and 12^{°} C for T_{i} and 2.40 kW for P in the equation t =[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)] (100°C -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 t for the kettle to boil dry will be,
= ρ_{w}V_{w}L_{w}/P
Here density of water is ρ_{w}, volume of water is V_{w}, latent heat of water is L_{w} and power is P.
To obtain the time t for the kettle to boil dry, substitute 998 kg/m^{3} for ρ_{w}, 0.640 L for V_{w}, 2256×10^{3} J/kg for L_{w} and 2.40 kW for P in the equation t = 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.
To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
THIRD LAW OF THERMODYNAMICS:- In all heat engines,...
Solved Examples on Thermodynamics:- Problem 1 :- A...
Enthalpy of Reaction It is the enthalpy change...
Enthalpy of a System The quantity U + PV is known...
Thermodynamic change or Thermodynamic Process:-...
HESS’S LAW This law states that the amount...
Level 2 Objective Problems of Thermodynamics Level...
Work Done During Isothermal Expansion:- Consider...
Introduction to Thermodynamics:- Thermodynamics:-...
Application of Hess's Law 1. Calculation of...
Relationship between free Energy and Equilibrium...
The First Law of Thermodynamics:- The first law of...
Objective Questions of Thermodynamics and Answers...
Work Done During Adiabatic Expansion:- Consider...
Macroscopic Properties He properties associated...
Solved Problems on Specific Heat, Latent Heat and...
Solved Problems on Thermodynamics:- Problem 1:- A...
Thermodynamic State of a System and Macroscopic...
Second Law of Thermodynamics:- Entropy:- The...
Specific Heat Capacity and Its Relation with...
Application of bond energies (i) Determination of...
Miscellaneous Exercises Thermal Physics:- Problem:...
Gibbs Free Energy This is another thermodynamic...
Reversible and Irreversible Process:- Reversible...
BOMB CALORIMETER The bomb calorimeter used for...