Heat transfer is a process is known as the exchange of heat from a high-temperature body to a low-temperature body. As we know heat is a kinetic energy parameter, included by the particles in the given system. As a system temperature increases the kinetic energy of the particle in the system also increases. The energy of the particle from the one system to other system is transferred when these systems are brought into contact with one another.

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

Where

Q = Heat supplied to the system

m = mass of the system

c = Specific heat capacity of the system and

Î”T = Change in temperature of the system.

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

- Â Â Â Conduction
- Â Â Â Convection
- Â Â Â Radiation.

**Conduction: **

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

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

Q = Heat transferred

K = Thermal conductivity

T_{HOT} = Hot temperature

T_{COLD} = Cold Temperature

t = Time

A = Area of the surface

d = Thickness of the material

**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 Ïƒ 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.

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

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

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 40 ^{o}C and the other being at 20^{o}C. 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 1.4 W/mK)**

**Solution:**

According to question,

Thermal Conductivity of glass = 1.4 W/mK.

Also, the temperature of the first column is T_{h}=40^{0} C andÂ

The temperature of the second column is T_{c}=20^{0} C.

Area of the wall separating both the columns = 1mÂ Ã—Â 2m = 2 m^{2}

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 30áµ’C to its final temperature of 60áµ’C. Calculated the total heat gained by the system. (Specific heat of the system = 0.45 kJ/Kg K)**

**Solution:**

According to question,

The Initial temperature of the system, TiÂ = 30áµ’C

The Final temperature of the system, TfÂ = 60áµ’C

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,

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

Q=5Ã—0.45Ã—303

**Q=681.75 J**