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

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