Q = mc ΔT

- Q amount of heat in J
- m mass in kg
- c specific heat capacity in Jkg
^{-1o}C^{-1}or Jkg^{-1}K^{-1} - ΔT change in temperature in
^{o}C of K

The specific heat capacity of water : 4.18 x 10^{3} Jkg^{-1}K^{-1}

**Example 1**

500 g water is heated in a pan. The temperature increases of 20^{o}C to 50 ^{o}C.

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 10^{3})(50-20) = 6.27 x 10^{4} J

(at a temperature drop of 30 ^{o} C, an energy of 6.27 x 10^{4} 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 ^{ o}C^{-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 10^{4} + 40 (50-20) = 6.27 x 10^{4} + 0.12 x 10^{4} = 6.39 x 10^{4 }J