Category Archives: Mahabharata

Heat Transfer Formula

Heat transfer is defined as the process of transfer of heat from a body at higher temperature to another body at a lower temperature. Heat, as we know, is the measure of kinetic energy possessed by the particles in a given system. When the temperature of a system is increased, the kinetic energy possessed by particles in the system increases. When this system is brought in contact with another system whose temperature is lesser as compared to the first, the energy gets transferred from the particles in the first system to that in the second system.

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

Q= m\times c\times \Delta T

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 ΔT is the change in temperature of the system.

The transfer of heat occurs through three different processes, which are mentioned below.

  1.    Conduction
  2.    Convection
  3.    Radiation.

Conduction:

Heat transferred by the process of conduction can be expressed by the following equation,

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

Convection:

Heat transferred by the process of convection can be expressed by the following equation,

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

Here, Hc is the heat transfer coefficient.

Radiation:

The Heat transferred by the process of radiation can be given by the following expression,

Q= \sigma \left ( T_{4}^{Hot}-T_{4}^{Cold} \right )A

Here σ is known as Stefan Boltzmann Constant.

Derivation:

From the definition of specific heat capacity, we can say that, it is the total amount of heat that is to be supplied to a unit mass of the system, so as to increase its temperature by 1 degree Celsius.

Let us consider a system of mass ‘m’. In order to calculate the relation between the rise in temperature with the amount of heat supplied, we multiply the specific heat of the system by the mass of the system and the desired rise in temperature.

Now, the total heat to be supplied to the system can be given as,

Q= m\times c\times \Delta T

Real Life Example: Let us consider a pitcher of water that is to be heated till its temperature rises from the room temperature to 100 degree Celsius. In this case, as we know the mass of the water and its specific heat capacity at the given conditions, we can use the above mentioned formula to calculate the amount of heat to be supplied.

Example 1

Let us consider two water columns at different temperatures, one being at 40oC and the other being at 20oC. As both the water columns are separated by a glass wall of area 1m by 2m and a thickness of 0.003m. Calculate the amount of heat transfer. (Thermal Conductivity of glass is 1.5 W/mK)

Solution:

According to question,

Thermal Conductivity of glass = 1.4 W/mK.

Also, the temperature of the first column is Th=40 C and the temperature of the second column is Tc=40 C.

Area of the wall separating both the columns = 1m × 2m = 2 m2

Using the heat transfer equation for conduction, we can write,

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

Q= \frac{1.4\times 2\times 20}{0.003}= 18667 W

Example 2

A system weighing 5 Kgs is heated from its initial temperature of 30ᵒC to its final temperature of 60ᵒC. Calculated the total heat gained by the system. (Specific heat of the system = 0.45 kJ/Kg K)

Solution:

According to question,

Initial temperature of the system, Ti = 30ᵒC,

Final temperature of the system, Tf = 60ᵒC,

Mass of the system, m = 5 kg,

The total heat gained by the system can be calculated by using the formula for heat transfer as mentioned above,

Q= m\times c\times \Delta T

Q=5×0.45×303=681.75 J