Content Curator
Specific Heat Capacity- Take equal quantities of water and oil at a temperature of about 30°C. Heat both of them till they reach the temperature 50°C. We can observe that water takes more time than oil to reach a temperature of 50°C. It implies that water requires more heat energy to raise its temperature than oil. From the observations, we can define specific heat capacity as the amount of energy required to raise the temperature of a given body
Specific Heat Capacity
Specific Heat Capacity is defined as the amount of heat energy required to raise the temperature of 1kg of a substance by 1 Kelvin or 1 °C.
We know that-
ΔQ = mSΔT
Therefore, Specific Heat Capacity can be expressed as:
S = ΔQ/ mΔT
Where,
- S is known as the Specific Heat Capacity
- ΔQ is the amount of heat energy
- m is the mass of a substance
- ΔT is a change in temperature
Heat capacity and Specific Heat capacity are always positive quantities. The SI unit of Specific Heat capacity is J kg−1K−1. SI units used for specific heat capacity in degree Celcius are J/kg°C and J/g°C.
The Specific Heat Capacity of a substance depends on three factors of a substance.
- Mass of the substance
- Change in temperature of the substance
- Nature of material of the substance
Specific Heat Capacity Video Explanation
Read More: Specific Heat Capacity in Detail
Heat Capacity and Specific Heat Capacity
The Heat Capacity represents the change in temperature in the sample for a given amount of heat. The SI Unit of Heat Capacity is joule per kelvin (JK-1).
Specific Heat Capacity represents the amount of heat needed to raise the temperature of a unit mass of a substance by 1°C. The SI units of Specific Heat Capacity are J kg−1K−1 and J/kg°C.
The term heat capacity or specific heat capacity does not mean that object contains a certain amount of heat. The transfer of energy from the object at a higher temperature to the object at a lower temperature is called heat. The correct usage is internal energy capacity. But for some reason, the terms “heat capacity” and “specific heat capacity” are retained.
However, heat capacity and Specific Heat capacity are different from each other in an aspect. Specific Heat capacity is related to the mass of a body, hence it is more accurate than the heat capacity.
Importance of Specific Heat Capacity
It is important to know the information regarding how long the heating or cooling process will take for a given substance. The amount of energy used, time taken, costs involved, and temperature change in the substances can be studied by calculating the specific heat capacity.
For Example:
The specific heat capacity of water is about 4200 J/kg.K
This means that it will take 4200 J of energy to raise the temperature of 1 kg of water by 1-degree kelvin.
So, if you want to boil 2 kg of water in an electric kettle from room temperature of 20°C to 100°C. Then, the energy required for the change will be,
Q= mcΔT
=2×4200×80
=672kJ
Suppose the kettle has a power rating of 2KW. Then the time taken to boil the water will be-
t=WP
=336s
=5.6mins
Suppose the power provider charges tariffs of Rs 1.50/Kwh unit.
So, the total cost to boil the water will be = Rs 0.28.
This is how specific heat capacity can be used to calculate the
Specific Heat capacity of some substances at Atmospheric Temperature (20°C) :
| Material | Specific Heat capacity in J kg−1K−1 |
| Air | 1005 |
| Lead | 130 |
| Copper | 390 |
| Iron | 450 |
| Glass | 840 |
| Aluminum | 900 |
| Human Body | 3470 |
| Water | 4200 |
- When two objects at the same mass are heated at equal rates, the temperature of the object with a smaller specific heat capacity will increase faster.
- When two objects of the same mass are cooled down at an equal rate, the temperature of the object with a smaller Specific Heat capacity will drop faster.
Molar Specific Heat Capacity
When estimating the properties of gases, it is more practical to use molar-specific heat capacity. Considering the point of moles instead of mass for the gases is more preferable.
Molar Specific Heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance.
In other words, molar Specific Heat capacity is the amount of heat required to raise the temperature of one gram molecule of substance by 1°C.
The SI unit of Molar specific heat capacity is J mol-1K-1.
Molar Specific Heat capacity can be expressed as-
C = S / μ
C = Q / (μ ΔT)
?Q = µ C ?T.
Here,
- C is Molar specific heat capacity
- ?Q is the amount of heat energy
- ΔT is a change in temperature
- μ is moles of solid.
Types of Molar Specific Heat Capacity
When the substances are heated, we only consider the amount of energy to be added or removed and do not specify anything about the phase change. It is because there is no change in temperature during the phase change.
Molar Specific Heat capacity is further divided into two types :
- Molar specific heat capacity at Constant Pressure: When a solid substance is heated at constant pressure, the specific heat obtained from such heat transfer is called Molar specific heat capacity at Constant Pressure. It is denoted by Cp.
- Molar specific heat capacity at Constant Volume: When a solid substance is heated at constant volume, the specific heat obtained from such heat transfer is called Molar specific heat capacity at Constant Volume. It is denoted by Cv.
Due to various factors and conditions, specific heat has infinite values. A constant heat supply must be provided to the gas in order to obtain a constant specific heat value. So the conditions of volume, evaporation, and pressure must be considered.
For an ideal gas:
Cp – Cv = nR
Where,
Cp is Molar specific heat capacity at constant Pressure
Cv is Molar specific heat capacity at constant volume
n is the amount of substance
R is molar gas constant = 8.3144598
Molar specific heat capacities of some gases :
| Gas | Cp in J mol-1K-1 | Cv in J mol-1K-1 |
| He | 20.8 | 12.5 |
| H2 | 28.8 | 20.4 |
| N2 | 29.1 | 20.8 |
| O2 | 29.4 | 21.1 |
| CO2 | 37 | 28.5 |
Specific Heat Capacity: Things to Remember
- Specific Heat Capacity is the amount of heat energy required to raise the temperature of 1kg of a substance by 1 Kelvin or 1 °C.
- Heat Capacity is defined as the change in temperature of a substance for a given amount of heat.
- Molar Specific Heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance.
- Molar specific heat capacity at Constant Pressure is the specific heat obtained by heating a solid substance Constant Pressure. It is denoted by Cp.
- Molar specific heat capacity at Constant Volume is the specific heat obtained by heating a solid substance at constant volume. It is denoted by Cv.
- The relation between Molar Specific Heat Capacity at Constant Volume and Molar Specific Heat Capacity at Constant Pressure can be given as Cp – Cv = nR.
Specific Heat Capacity: Important Questions
Ques. 1: A metal piece of 50 g specific heat 0.6 cal/g°C initially at 120°C is dropped in 1.6 kg of water at 25°C. Find the final temperature or mixture. (2 Marks)
Ans: Given that :
m1 = 50 g
C1 = 0.6 cal/gm?
m2 = 1.6 kg
C2 = 1 cal/gm?
At Equilibrium:
Heat lost by the body = Heat gained by the body
m1C1(Θ1 - Θ) = m2C2 (Θ - Θ2)
50 x 0.6 x (120 - Θ) = 1.6 x 1000 x 1x (Θ - 25)
Θ = 26.8 ?
Ques.2 : Calculate heat required to convert 3kg of water at 0°C to steam at 100°C Given specific heat capacity of H20 = 4186J kg-1 k-1 and latent heat of stream = 2.256 x 106 J/kg (2 Marks)
Ans: Given that:
Mass of Water m1 = 3 kg
Specific Heat Capacity c1 = 4186 J kg-1 k-1
Temperature t = 100 °C
Latent heat of steam = 2.256 x 106 J/kg
Heat required to convert H2O at 0°C to H2O at 100 °C = m1c1 t
= 3 × 4186 × 100
= 1255800 J
Heat required to convert H2O at 100°C to steam at 100°C is = mL
= 3 × 2.256 x 106
= 6768000J
Total Heat = 80.23800 J
Ques.3: A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg-1K-1; heat of fusion of water = 335 Jg-1). (2 Marks)
Ans: Mass of the copper block, m = 2.5 kg = 2500 g
Rise in the temperature of the copper block, Δθ = 500°C
Specific heat of copper, C = 0.39 Jg-1K-1
Heat of fusion of water, L = 335 Jg-1
The maximum heat the copper block can lose, Q = mCΔθ
= 2500 x 0.39 x 500
= 487500 J
Let m1g be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, Q = m1L
m1 = Q / L = 487500 / 335 = 1455.22 g
Hence, the maximum amount of ice required to melt the ice is 1.45 kg.
Ques.4: The Coolant used in a nuclear plant should have high specific heat. Why? (2 Marks)
Ans: The Coolant used in a nuclear plant should have high specific heat because it absorbs more heat and shows a very small change in temperature. Due to this it also extracts a large amount of heat.
Ques.5: The specific heat of gold is 0.129 J/g degrees Celsius. What is the molar heat capacity of gold? (2 Marks)
Ans: Specific heat of gold, c =0.129 J/g°C
To calculate the molar heat capacity of gold,
cm= c×M
Here, M =molar mass of the gold = 196.96 g/mol
cm=c×M
cm=0.129 × 196.96
=0.129 × 196.96
=25.4 J/mol°C
Ques.6: A sample of ideal gas (γ=1.4) is heated at constant pressure. If an amount of heat 140 J is supplied to the gas, find the work done by the gas? (2 Marks)
Ans: Let n be the no of moles of the gas.
Let the temperature change from T1 to T2 and volume from V1 to V2 respectively.
So, the total heat supplied
Q = nCp(T2−T1) = 140
Change in internal energy:
ΔU = nCv(T2−T1) = nγCp(T2−T1)
=1.4140
=100J
So, the work done by the gas = ΔQ−ΔW
=140−100
=40J
Ques.7: A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg-1K-1. (3 Marks)
Ans: Given that:
Power used for drilling machine, P= 10 kW = 10 x 103 W
Mass of the aluminum block, m= 8.0 kg = 8 x 103 g
Time taken by the machine, t= 2.5 min = 2.5 x 60 = 150 s
Specific heat of aluminium, c = 0.91 Jg-1K-1
Rise in the temperature of the block after drilling = δ T
Total energy required by the drilling machine = Pt
= 10 x 103 X 150
= 1.5 x 106 J
It is given that only 50% of the power is useful.
Energy used, ?Q = 50/100 x 1.5 x 106 J = 7.5 x 105
But ?Q = m C ?T
?T = ?Q / mc
= 7.5 x 105 / 8 x 103 x 0.91
= 103 °C
Therefore, in 2.5 minutes the temperature of the block will rise up to 103°C.
Ques.8: In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal? (3 Marks)
Ans: Given that:
Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T1= 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m' = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C:
150 x 1 = 150 g
Fall in the temperature of the metal:
ΔT =T1 -T2 = 150 - 40 = 110°C
Specific heat of water, Cw= 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT … (i)
Rise in the temperature of the water and calorimeter system:
ΔT’ = 40-27 = 13°C
Heat gained by the water and calorimeter system:
Δθ’’ = m1CwΔT’
= (M + m’)CwΔT’ … (ii)
Heat lost by the metal = Heat gained by the water and calorimeter system
mCΔT = (M + m') CwΔT'
200 x C x 110 = (150 + 25) x 4.186 x 13
C = 175 x 4.186 x 13 / 110x200 = 0.43 Jg-1K-1
The value of C will be smaller than the actual volume if there will be some lost in the heat.
Ques.9: Given below are observations on molar specific heats at room temperature of some common gases.
| Gas | Molar specific heat (Cv) (cal mol K) |
| Hydrogen | 4.87 |
| Nitrogen | 4.97 |
| Oxygen | 5.02 |
| Nitric oxide | 4.99 |
| Carbon monoxide | 5.01 |
| Chlorine | 6.17 |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine? (3 Marks)
Ans: The gases mentioned in the table are diatomic in nature. Besides the translational degree of freedom, all these gases have other degrees of freedom.
Some amount of heat must be supplied to increase the temperature of these gases. The heat supplied will increase the average energy of all the degrees of freedom. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.
If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = 5/2 R
5/2 x 1.98 = 4.95 cal mol-1K-1
With the exception of chlorine, all the observations in the given table agree with 5/2 R. This is because at room temperature, chlorine also has vibrational degrees of freedom besides rotational and translational degrees of freedom.
Ques.10: A child running a temperature of 101°F is given an antipyrine (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of the human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g-1. (5 Marks)
Ans: Given that:
Initial temperature of the body of the child, T1= 101°F
Final temperature of the body of the child, T2= 98°F
Change in temperature, ΔT = (101-98) x 5/9 °C
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 x 103 g
Specific heat of the human body = Specific heat of water = c = 1000 cal/kg/oC
Latent heat of evaporation of water, L = 580 calg-1
The heat lost by the child can be given as-
Δθ = m1CΔT
= 30 x 1000 x (101-98) 5/9
= 50000 cal
Let m1 be the mass of the water evaporated from the child's body in 20 min.
Loss of heat through water is given by:
Δθ = m1L
m1 = Δθ / L
=50000 / 580 = 86.2g
∴Average rate of extra evaporation caused by the drug = m1/ t = 4.3 g min-1
Comments