Heat Transfer Formulas

Heat is a measure of thermal energy that can be transferred from one point to another. Heat is the transfer of kinetic energy from an energy source to a medium or from one medium or object to another medium or object.

Heat is one of the important components of phase changes associated with work and energy. Heat is also the measure of kinetic energy possessed by the particles in a system. The kinetic energy of the particles in the system increases with the increase in the temperature of the system. Hence, heat measures changes with time.

Heat Transfer

When a system at a higher temperature is brought in contact with a system at a lower temperature, energy is transferred from the particles in the first system to the particles in the second.

Therefore, heat transfer can be defined as the process of the transfer of heat from an object (or a system) at a higher temperature to another object (or a system) at a lower temperature. 

The heat transfer formula determines the amount of heat transferred from one system to another.

Q = c × m × ΔT

Where,

Q is the heat supplied to the system

m is the mass of the system

c is the specific heat capacity of the system 

ΔT is the change in temperature of the system

The specific heat capacity (c) is defined as the quantity of heat (in Joules) absorbed per unit mass (kg) of the material when its temperature increases by 1 K (or 1 °C). Its units are J/kg/K or J/kg/°C.

Derivation of the Formula,

Let 'm' be the mass of the system and 'c' be the specific heat capacity of the system. Let ΔT be the change in temperature of the system.

Then the amount of heat supplied (Q) is the product of the mass m, specific heat capacity 'c' and change in temperature ΔT and is given by,

Q = c × m × ΔT

Types of Heat Transfer

There are three types of heat transfer:

1. Conduction
2. Convection
3. Radiation

Conduction

The transfer of heat through solid materials is called conduction. The formula for heat transferred by the process of conduction is expressed as:

Q = kA(T_Hot-T_Cold)t/d

Where,  

Q is heat transferred through conduction

k is thermal conductivity of the material

A is the area of the surface

T_Hot is the temperature of the hot surface

T_Cold is the temperature of the cold surface

t is time

d is the thickness of the material

Convection

The transfer of heat through liquids and gases is called convection. The formula for heat transferred by the process of convection is expressed as:

Q = H_cA(T_Hot-T_Cold)

Where, 

Q is heat transferred through convection

H_c is the heat transfer coefficient 

A is the area of the surface

T_Hot is the temperature of the hot system

T_Cold is the temperature of the cold system

Radiation

The transfer of heat through electromagnetic waves is called radiation. The formula for heat transferred by the process of radiation is expressed as:

Q = σ (T_Hot - T_Cold)⁴A

Where, 

Q is heat transferred through radiation

σ is Stefan Boltzmann Constant

T_Hot is the temperature of the hot system

T_Cold is the temperature of the cold system

A is the area of the surface

Stefan Boltzmann Constant (σ) is calculated as:

σ = 2.π⁵ K_B⁴ / 15 h³ c² = 5.670367(13) × 10^-8 J . m^-2. S^-1 . K^-4

Where,

σ is Stefan Boltzmann Constant

pi(π) ∼= 3.14

k_B is Boltzmann constant

h is Planck's constant

c is speed of light in vacuum

Sample Problems

Problem 1: A system with a mass of 10 kg and an initial temperature of 200 K is heated to 450 K. Specific heat capacity of the system is 0.91 KJ/kg K. Calculate the heat gained by the system in this process.

Solution:

According to question,

Mass, m = 10 kg

Specific heat capacity, c = 0.91 KJ/kg K

Initial temperature, T_i = 200 K

Final temperature, T_f = 450 K

Change in temperature, ΔT = 450K - 200K = 250K

Using the heat transfer formula,

Q = c × m × ΔT

Q = 0.91 x 10 x 250

Q = 2275 KJ

Therefore the total heat gained by the system is 2275 KJ.

Problem 2: The specific heat of iron is 0.45 J/g°C. What mass of iron is required for a heat transfer of 1200 Joules if the temperature change is 40°C?

Solution:

According to question,

Specific heat of iron, c = 0.45 J/g°C

Change in temperature, ΔT = 40°C

Amount of heat transferred, Q = 1200 J

Using the heat transfer formula,

Q = c × m × ΔT

m = Q /(c x ΔT)

m = 1200 /(0.45 x 40)

m = 66.667 g

Therefore required mass of iron for a heat transfer of 1200 Joules is 66.667 grams.

Problem 3: Consider two water columns at different temperatures separated by a glass wall of length 3m and width 1.5m and a thickness of 0.005m. One water column is at 380K and the other is at 120K.  Calculate the amount of heat transferred if the thermal conductivity of glass is 1.4 W/mK.

Solution:

According to question,

Thermal Conductivity of glass, k = 1.4 W/mK.

Temperature of first water column, T_{Hot= 380K}

Temperature of second water column, T_{Cold = 120K}

Area of the glass wall separating two columns, A = length x width = 3m x 1.5m = 4.5m²

Thickness of the glass, d = 0.005m

Using the heat transfer formula for conduction, 

Q = kA(T_Hot-T_Cold)t / d

Q = 1.4 x 4.5 (380-120) / 0.005

Q = 327600 W

Therefore, amount of heat transferred is 327600 Watts.

Problem 4: Calculate heat transfer through convection if the heat transfer coefficient of a medium is 8 W/(m^2​ K) and thearea is 25 m² and the temperature difference is 20K.

Solution:

According to question,

Heat transfer coefficient, H_c = 8 W/(m^2​ K)

Area, A = 25m²

Change in temperature, (T_Hot - T_Cold) = 20K

Using the heat transfer formula for convection,

Q = H_cA(T_Hot-T_Cold)

Q = 8 x 25 x 20

Q = 4000 W

Therefore, amount of heat transferred through convection is 4000 Watts.

Problem 5: Calculate the heat transferred through radiation between two black bodies at temperatures 300K and 430K and the area of the medium is 48 m². (Given Stefan Boltzmann Constant, σ = 5.67 x 10^-8 W/(m²K⁴) ).

Solution:

The formula for heat transfer through radiation between two black bodies is

Q=σ⋅A⋅(T⁴Hot4​−T⁴Cold)

Where,

σ is the Stefan-Boltzmann constant, σ=5.67×10−8 W/m²K⁴

A is the area through which heat is transferred, A=48 m²

T_Hot and T_Cold​ are the temperatures of the hot and cold bodies, respectively

Given:

T_Hot=430 K

T_Cold=300 K

A=48 m²

σ=5.67×10⁻⁸ W/m²K⁴

Now, substitute the values:

Q=5.67×10⁻⁸×48×(430⁴−300⁴)

Let's calculate the values:

T⁴Hot=430⁴=3.419×10⁹

Now substituting back :

Q=5.67×10−⁸×48×3.338×10⁹

Q=5.67×10−⁸×1.599×10¹¹

Q=907.6W