Heat Transfer Formula ,some problems and solution

  Heat Transfer Formula

                                Heat transfer is all about the transfer of heat from one point to another. If we consider any system which will be at higher temperature compared to surroundings, there will be transfer of heat from system to the surroundings.

                            Heat transfer is given by

Where,   m is the mass,
             C is the specific heat and
             ΔT is the temperature difference in K.


There are three types of Heat transfer

1. Conduction
2. Convection
3. Radiation.

                         Heat transfer (Q) by Conduction formula is given by

Where, k is the thermal conductivity of the material,
           A is the cross sectional area,
           T_Hot is the higher temperature,
           T_Cold is the cooler temperature,
           t is the time taken,
           d is the thickness of the material.

                             Heat transfer by Convection is given by
                 Where H_c is the heat transfer coefficient,

                        The Heat transfer by radiation is given by
              Here σ is the Stefan Boltzmann Constant.

                 Heat transfer is usually expressed in Joules. Heat transfer formula is used in problems to find the heat transfer taking place for any given material.

                  

                               Heat Transfer Problems

Below are some problems on Heat transfer which may be helpful for you.

Solved Examples

Question 1: A room is maintained at 20^oC where as temperature outside the room is 10^oC. There is one of the window in the corner having area 1m by 2m. Calculate the heat transfer?
Thermal Conductivity of glass is 1.4 W/mK.
Solution:

Thermal Conductivity of glass is 1.4 W/mK.
Initial temperature Tcold = 10^oC,
Final temperature THot = 20^oC,
Area A = 1m ×$\times$ 2m = 2 m²,

The Heat transfer Q = kA(THot−TCold)d$\frac{kA(T_{Hot}-T_{Cold})}{d}$
                             = 1.4W/mK×2m2×10oC0.003m$\frac{1.4W/mK\times 2m^2\times {10}^{o}C}{0.003m}$
                   

                           = 9333.33 W.

Question 2: Calculate the Heat lost by the block when iron block decreases its temperature from 60^oC to 40^oC if the mass of the body is 2 Kg.
Specific heat of iron C = 0.45 kJ/kg K.
Solution:

Given: Initial temperature T_i = 60^oC,
          Final temperature T_f = 40^oC,
          Mass of the body m = 2 kg,
The Heat lost is given by Q = m c Δ$\mathrm{\Delta }$ T
                                        = 2 Kg ×$\times$ 0.45 kJ/kg K ×$\times$ 293 K
                                        = 263.7 J.