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