Concept of Specific Heat and latent Heat

Science > Physics > Thermometry > Numerical Problems on Temperature Scales

In the previous article, we studied measurement temperature, various scales to measure temperature viz: Celsius scale, Fahrenheit scale, and Kelvin scale, and also interconversion of temperature in different temperature scales. In this article, we shall study, the concept of specific heat and latent heat and numerical problems on them.

Concept of Heat:

A body is made up of a large number of particles, called molecules. Due to the kinetic energy of these constituent molecules by virtue of their motion viz: translational, vibrational, rotational, the body possesses heat or thermal energy.

When a hot body is placed in a contact with a cold body, heat is transferred from the hot body to the cold body. This transfer of energy between two bodies as a result of temperature difference is called heat. Thus heat is the energy that is transferred from one body to another due to the temperature difference between the two bodies.

S.I. unit of heat is joule (J), c.g.s. unit is calorie (cal).

• The amount of heat required to raise the temperature of 1 kg of water through 1 K is called 1 joule.
• The amount of heat required to raise the temperature of 1 gram of water through 1^oC is called 1 calorie.
• Relation between calorie and joule: 1 calorie = 4.186 J

Specific Heat or Heat Capacity:

When heat is supplied to a solid (or a liquid) substance, its temperature increases. It is found that amount of heat Q absorbed by a solid (or a liquid) substance is

• directly proportional to the mass (m) of the substance

i.e. Q α m  …… (1)

• directly proportional to rise in temperature

i.e. Q α ΔT …… (2)

Thus, Q α mΔT

Q = c m ΔT

Where ‘c’ is a constant of proportionality and is called the specific heat of the substance.

Hence, C = Q/m Δ T

Let m = 1 kg and Δ T = 1 K

C = Q

Hence specific heat of a solid (or a liquid) substance is defined as the amount of heat required to raise the temperature of 1 kg of the solid (or the liquid) through 1  K (or 1oC). S.I. Unit of C is J/kg/K (J kg^-1K^-1). Specific heat of a substance is also known as its heat capacity.

The molar specific heat of solid (or liquid) is defined as the amount of heat required to raise the temperature of 1 mole of solid (or liquid) through 1 K (or 1^oC). It is denoted by C

Q = c n ΔT

Where n = number of moles of solid (or lquid)

C = Mc

Where, C = Molar specific heat, M = Molecular mass of sold (or liquid), and c = Specific heat of a substance.

The specific heat of water is 1 cal/g/^oC or 4200 J/kg/K.

Change of Phase and Latent heat

The phase of a substance is defined as its form which is homogeneous, physically distinct, and mechanically separable from other forms of the substance. The term phase as used in thermodynamics refers to the fact that the matter exists either as a solid, liquid, or gas. If we consider the example of water, it exists in the solid phase as ice, in the liquid phase as water, and in the gaseous phase as vapour. All the substances can exist in any of the three phases under proper conditions of temperature and pressure.

Change of Phase:

The transitions from one phase to another take place by the absorption or liberation of heat and usually by a change in volume and at a constant temperature. The temperature at which a phase change occurs also depends on pressure.

Latent Heat:

The heat absorbed or released by a substance during the change of its physical state at constant temperature is called the latent heat of the substance for that physical change.

Latent Heat of Fusion:

The quantity of heat required to convert the unit mass of a solid into a liquid state completely at its melting point and at constant temperature is called the latent heat of fusion.

The latent heat of the fusion of water is 334 kJ/kg, which means that to change 1 kg of ice at 0^oC into liquid water at 0^oC completely, we must supply 334 kJ of heat must be supplied.

Latent Heat of Vapourization:

The quantity of heat required to convert the unit mass of a liquid into a gaseous state completely at its boiling point and at constant temperature is called the latent heat of vaporization.

The latent heat of the vapourization of water is 2257 kJ/kg, which means that to change 1 kg of water at 100^oC into steam at 100^oC completely, we must supply 2257 kJ of heat must be supplied.

Numerical Problems on Specific and Latent Heats:

Example 01:

What is the specific heat of a metallic substance if 36 kcal of heat is needed to raise 4.5 kg of metal from 20 ^oC to 42 ^oC?

Given: Mass of substance = m = 4.5 kg = 4.5 x 10³ g, Initial temperature = θ₁ = 20 ^oC, Final temperature = θ₂ = 42 ^oC, amount of heat needed = Q = 36 Kcal = 36 x 10³ cal

To Find: specific heat of metal = c =?

Solution:

Q = m c (θ₂ – θ₁)

∴ 36 x 10³ = 4.5 x 10³ x c x (42 – 20)

∴ 36 = 4.5 x c x 22

∴ c = 36/(4.5 x 22) = 0.3636 = 0.36 cal/g/^oC

Ans: Specific heat of metallic substance is 0.36 cal/g/^oC

Example 02:

60 cal of heat is added to 15 g of water, find rise in the temperature of water.

Given: Mass of water = m = 15 g, Amount of heat added = Q = 60 cal, Specific heat of water = c = 1 cal/g/^oC

To Find: Rise in temperature = (θ₂ – θ₁) =?

Solution:

Q = m c (θ₂ – θ₁)

∴ 60 = 15 x 1 x (θ₂ – θ₁)

∴ (θ₂ – θ₁) = 60/15 = 4 ^oC

Ans: Rise in temperature of water is 4^oC

Example 03:

A piece of lead gives out 1200 cal of heat when it it is cooled from 90 ^oC to 10 ^oC. What is its water equivalent?

Given: Initial temperature = θ₁ = 90 ^oC, Final temperature = θ₂ = 10 ^oC, amount of heat given out = Q = 1200 cal

To Find: water equivalent of lead = mc =?

Solution:

Q = m c (θ₂ – θ₁)

∴ 1200 = m c x (90 – 10)

∴ 1200 = 80 mc

∴ m c = 1200/80 = 15 g

Ans: The water equivalent of lead is 15 g

Example 04:

Calculate the amount of heat required to heat 10 g of solid from -10 ^oC to 10 ^oC, its specific heat being 0.1 cal/g/^oC

Given: Mass of solid = m = 10 g, Initial temperature = θ₁ = – 10 ^oC, Final temperature = θ₂ = 10 ^oC, specific heat of solid = 0.1 cal/g/ ^oC

To Find: Amount of heat required = Q =?

Solution:

Q = m c (θ₂ – θ₁)

∴ Q = 10 x 0.1 x (10 – (-10))

∴ Q = 10 x 0.1 x 20

∴ Q = 20 cal

Ans: Amount of heat required = 20 cal

Example 05:

Calculate the heat required to convert 10 kg of ice at – 20^oC to steam 100 ^oC at normal atmospheric pressure. Specific heat capacity of ice = 2100 J/kg/K and of water 4186 J/kg/K. Latent heat of fusion of ice = 3.35  J/kg and steam of = 2.26  J/kg.

Solution:

Q₁ = Heat required to convert 10 kg (m) of ice at – 20 ^oC ( θ₁) to ice at 0 ^oC ( θ₂)

∴ Q₁ = m C Δ θ = m c_ice ( θ₂ – θ₁) = 10 x 2100 x (0 – (-20)) = 420000 J = 0.42 x 10⁶ J

Q₂ = Heat required to convert 10 kg (m) of ice at 0 ^oC to water at 0 ^oC

∴ Q₂ = m L = m L_fusion = 10 x 3.35 x 10⁵ = 3.35 x 10⁶ J

Q₃ = Heat required to convert 10 kg (m) of water at 0 ^oC ( θ ₁) to water at 100 ^oC ( θ ₂)

∴ Q₃ = m c Δ θ = m c_water ( θ₂ – θ₁) = 10 x 4186 x (100 – 0) = 4.186 x 10⁶ J

Q₄ = Heat required to convert 10 kg (m) of water at 100 ^oC to steam at 100 ^oC

∴ Q₄ = m L = m L_{vapourization} = 10 x 2.26 x 10⁶ = 22.6 x 10⁶ J

∴ Total heat required = Q = Q₁ + Q₂ + Q₃ + Q₄

∴ Q = 0.42 x 10⁶ + 3.35 x 10⁶ + 4.186 x 10⁶ + 22.6 x 10⁶ = 30.556 x 10⁶ J = 30556 kJ

Ans: The heat required to convert 10 kg of ice at – 20^oC to steam 100 ^oC at normal atmosphere pressure = 30.556 x 10⁶ J = 30556 kJ

Example 06:

How much heat would be required to evaporate 500 g of ice at -5 ^oC? Sp.ht of ice = 0.5; L of fusion of ice = 80 cal/g; L of vaporization of water = 540 cal/g.

Solution:

Q₁ = Heat required to convert 500 g (m) of ice at – 5 ^oC ( θ ₁) to ice at 0 ^oC ( θ ₂)

∴ Q₁ = m c Δ θ = m c_ice ( θ ₂ – θ ₁) = 500 x 0.5 x (0 – (-5)) = 1250 cal = 1.250 x 10³ cal

Q₂ = Heat required to convert 500 g (m) of ice at 0 ^oC to water at 0 ^oC

∴ Q₂ = m L = m L_fusion = 500 x 80 = 40000 = 40 x 10³ cal

Q₃ = Heat required to convert 500 g (m) of water at 0 ^oC ( θ ₁) to water at 100 ^oC ( θ ₂)

∴ Q₃ = m c Δ θ = m C_water ( θ ₂ – θ ₁) = 500 x 1 x (100 – 0) = 50000 = 50 x 10³ cal

Q₄ = Heat required to convert 500 g (m) of water at 100 ^oC to steam at 100 ^oC

∴ Q₄ = m L = m L_{vapourization} = 500 x 540 = 270000 cal = 270 x 10³ cal

∴ Total heat required = Q = Q₁ + Q₂ + Q₃ + Q₄

∴ Q = 1.250 x 10³ + 40 x 10³ + 50 x 10³ + 270 x 10³ = 361.25 x 10³ cal = 361.25 Kcal

Ans: The heat required to convert 500 g of ice at – 5^oC to evaporate completely = 361.25 x 10³ cal = 361.25 Kcal

Example 07:

How much heat would be required to evaporate 600 g of water at 25 ^oC? L of vaporization of water = 540 cal/g.

Q₁ = Heat required to convert 600 g (m) of water at 25 ^oC ( θ₁) to water at 100 ^oC ( θ ₂)

∴ Q₁ = m c Δ θ = m c_water ( θ ₂ – θ ₁) = 600 x 1 x (100 – 25) = 45000 cal = 45 x 10³ cal

Q₂ = Heat required to convert 600 g (m) of water at 100 ^oC to steam at 100 ^oC

∴ Q₂ = m L = m L_{vapourization} = 600 x 540 = 324000 cal = 32.4 x 10³ cal

Total heat required = Q = Q₁ + Q₂

∴ Q = 45 x 10³ + 32.4 x 10³  = 77.4 x 10³ cal = 77.4 Kcal

Ans: The heat required to evaporate 600 g of water at 25 ^oC completely = 77.4 x 10³ cal = 77.4 Kcal

Example 08:

How much heat would be required to heat 500 g of water at 25 ^oC to 65 ^oC?

Solution:

Q = Heat required to convert 500 g (m) of water at 25 ^oC ( θ₁) to 65 ^oC ( θ₂)

∴ Q = m c Δ θ = m c_water ( θ₂ – θ₁) = 500 x 1 x (65 – 25) = 20000 cal = 20 Kcal

Ans: The heat required to heat 500 g of water at 25 ^oC to 65 ^oC is 20000 cal = 20 Kcal.

Example 09:

Calculate the amount of heat required to heat 10 g of a solid from – 10^oC to 10^oC; its specific heat being 0.1 cal/g/K.

Q = Heat required to convert 10 g (m) of solid at – 10 ^oC ( θ₁) to 10 ^oC ( θ₂)

Q = m c Δ θ = m c_water ( θ₂ – θ₁) = 10 x 0.1 x (10 – (-10)) = 20 cal Ans: Ans: The heat required to heat 10 g of a solid from – 10^oC to 10^oC is 20 cal.