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Physics Formulas

Heat Transfer Formula

Heat is an important component of phase changes related to work and energy. Heat transfer can be defined as the process of transfer of heat from an object at a higher temperature to another object at a lower temperature. Therefore heat is the measure of kinetic energy possessed by the particles in a given system. When the temperature of a system increases then its kinetic energy of the particles will also increase. Thus heat measure of an object changes with time due to many reasons. In this article, we will discuss the Heat Transfer Formula with examples. Let us begin the concept!

heat transfer formula

                                                                                                                                         Source: en.wikipedia.org 

Heat Transfer Formula

Concept of Heat Transfer:

When one system is brought in contact with the other system with low temperature then the energy gets transferred from the particles in the first system to the second system. So, there are many occasions when heat will be transferred from one object to the other one.

According to the definition of specific heat capacity, we can say that, the total amount of heat that has to be supplied to a unit mass of the system. Therefore, to increase the temperature by one degree Celsius.

The formula for Heat Transfer:

Let us consider a system of mass m Kg. In order to compute the relation between the rises in temperature with the amount of heat supplied, we have to multiply the specific heat of the system by the mass of the system and the rise in the temperature.

Heat transferred from one system to another system is given by the following equation,

\(Q=m \times c \times \Delta T\)

Where,

Q Heat transferred
m Mass
c Specific Heat
\(\Delta T\) Difference in temperature

Here, Q is the heat supplied to the system, m is the mass of the system, c is the specific heat capacity of the system and \(\Delta T\) is the change in temperature of the system.

The transfer of heat occurs through three different processes which are, Conduction, Convection, and Radiation.

Heat transferred by the process of conduction:

\(Q= \frac{kA\left ( T_{Hot}-T_{Cold} \right )t}{d}\)

Q Heat transferred
K Thermal Conductivity
\(T_{HOT}\) Hot temperature
\(T_{COLD}\) Cold Temperature
t Time
d The thickness of the material
A Area of surface

Heat transferred by the process of convection:

\(Q= H_{C}A\left ( T_{Hot}-T_{Cold} \right )\)

Where,

Q Heat transferred
\(H_{C}\) Heat Transfer Coefficient
\(T_{Hot}\) Hot temperature
\(T_{Cold}\) Cold Temperature
A Area of surface

The Heat transferred by the process of radiation:

\(Q= \sigma \left ( T_{Hot}-T_{Cold} \right )A\)

Where,

Q Heat transferred
\(\sigma\) Stefan Boltzmann Constant
\(T_{Hot}\) Hot temperature
\(T_{Cold}\) Cold Temperature
A Area of surface

Solved Examples

Q.1: A system of weight 7 Kg is heated from its initial temperature of \(30 ^{\circ} C\) to its final temperature of \(60^{\circ} C\). Find out the total heat obtained by the system. Note that the specific heat of the system is 0.45 kJ per Kg K.

Solution:

Initial temperature of the system, \(T_i = 30 ^{\circ} C\),

Final temperature of the system, \(T_f = 60^{\circ} C\),

Mass of the system, m = 7 kg,

The total heat gained by the system can be computed by using the formula for heat transfer as given:

\(Q = m \times c \times \Delta T\)

\(Q = m \times c \times (T_f – T_i) \)

so,

\(Q = 7 \times 0.45 \times (60 – 30) \)

\(Q = 7 \times 0.45 \times 30 \)

therefore,

\(Q = 94.5\; J\)

Therefore total heat obtained is 94.5 Joule.

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