FE Physics
 Formulas & Explanation
 
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Dutch version

Heat

           

Q = mc ΔT 

Q  amount of heat in J

m  mass in kg

c    specific heat capacity in Jkg-1oC-1 or Jkg-1K-1

ΔT change in temperature in oC of K  

The specific heat capacity of water : 4.18 x 103  Jkg-1K-1

 

Example 1

500 g water is heated in a pan. The temperature increases of 20oC to 50 oC.

The heat capacity of the pan is neglected.

Calculate the amount of energy needed to heat the water.

 

Q = mc ΔT 

Q = (0.500)( 4.18 x 103 ) (50-20) = 6.27 x 104  J

(at a temperature drop of 30 o C,  an energy of  6.27 x 104 J is released)

            

 C : heat capacity : the gained/lost energy per degree rise/drop in temperature               

C = Q/ΔT

also applies : C = mc.          

C  heat capacity  in JK-1 or in J  oC-1

 

Example 2

We start from the example above, but now the pan in which the water is heated  has a heat capacity of C = 40 JK-1

 

 Q = mc ΔT + C ΔT = 6.27 x 104 + 40 (50-20) = 6.27 x 104 + 0.12 x 104 = 6.39 x 104 J

 

Efficiency  (η)

 

η = Puseful/Pin     or   η = Euseful/Ein

 

η                 efficiency  (no dimension)

Puseful      useful power out in W

Pin              total power in in W

Euseful      useful energy out in J

Ein              total energy in in J

 

Example 3

In example 2 the water is heated by a an electric heating element with a power of  1000 W.

The water was heated for  1.5 minutes.

 

Calculate the efficiency of the process

 

Ein  = P  t = (1000) ( 1.5 x 60)  = 9.0 x 104  J

Euseful = Q = 6.39 x 104  J

η = (6.39 x 104)/(9.0 x 104) = 0.71  or  71 %