How much heat is required to convert 9.00 g of water at 100°c into steam at 100°c, if ∆hvap for water = 40.7 kj/mol?

There are two types of heat transfers in this process that are as follows:

q1= heat required to warm the water from 100 °C to 100.0 °C.

q2= heat required to vapourize the water into steam at 100 °C.

c=specific heat of water= 4.184 J g⁻¹ °c⁻¹

m=mass of water=9 g

∆Hvap=enthalpy of vapourisation

ΔT=temperature difference

q1=mcΔT=9  × 4.184 J ×0=0 J

q2=mΔHvap= 9 g × 0.0407 J=0.3663 J

q1+q2= (0 + 0.3663) J  

= 3.6 × 10⁻¹  J  

=  3.6 × 10⁻¹ J

16

explanation:

the marks go by 4s