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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
Formula for Heat Transfer
Where,m is the mass,
           C is the specific heat and
           $\Delta$ T 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,
           THot is the higher temperature,
           TCold is the cooler temperature,
           t is the time taken,
           d is the thickness of the material.

Heat transfer by Convection is given by
Heat Transfer by Convection FormulaWhere Hc is the heat transfer coefficient,

The Heat transfer by radiation is given by
Heat Transfer by Radiation FormulaHere $\sigma$ 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.

Related Calculators
Heat Transfer Calculator Specific Heat Formula Calculator
Heat Calculator Heat Capacity Calculator
 

Heat Transfer Problems

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

Solved Examples

Question 1: A room is maintained at 20oC where as temperature outside the room is 10oC. 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 = 10oC,
Final temperature THot = 20oC,
Area A = 1m $\times$ 2m = 2 m2,

The Heat transfer Q = $\frac{k A(T_{Hot} - T_{Cold})}{d}$
                             = $\frac{1.4 W/mK \times 2 m^{2} \times 10^{o} C}{0.003 m}$
                             = 9333.33 W.

 

Question 2: Calculate the Heat lost by the block when iron block decreases its temperature from 60oC to 40oC if the mass of the body is 2 Kg.
Specific heat of iron C = 0.45 kJ/kg K.
Solution:
 
Given: Initial temperature Ti = 60oC,
          Final temperature Tf = 40oC,
          Mass of the body m = 2 kg,
The Heat lost is given by Q = m c $\Delta$ T
                                        = 2 Kg $\times$ 0.45 kJ/kg K $\times$ 293 K
                                        = 263.7 J.
 

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