Question

How much thermal energy (in J) is required to boil 2.25 kg of water at 100.0°C into steam at 149.0°C? The latent heat of vaporization of water is 2.26 ✕ 10⁶ J/kg and the specific heat of steam is

2010

J

kg · °C

.

Answer 1

first 2.25kg of water at 100 degree celsius first become 2.25kg of steam at 100 degree celsius

heat required = mLf

                      = 2.25* 2.26*10^6

                    =5.085*10^6j

now 2.25 kg of steam at 100 degree celsius become 2.25 kg of steam at 149 degree celsius

heat required = mc (Tf-Ti)

                    = 2.25* 2010(49)

                    =.2*10^6j

total heat required = 5.285*10^6j