Answer
The amount of heat required for vaporization of 2.8 g of $H_{2}O$ is 6.8 kJ.
Work Step by Step
The amount of heat required to vaporize one mole of liquid is defined as a heat of vaporization.The heat of vaporization is represented by dH vap.
dH vap of water is 40.7 kJ/mole.
Given the mass of sweat is 2.8 g, and it vaporizes at 25c.
So, 1 mol of water = 18.02 g of water and dH vap of water = 44 kJ.
Hence conversion factors are :
(1mol $H_{2}O$/18.02 g $H_{2}O$) and (44 kJ/ 1mol $H_{2}O$).
Hence for dH vap of 2.8 g of $H_{2}O$ can be calculated as follows:
= (2.8 g $H_{2}O$) x(1mol $H_{2}O$/18.02 g $H_{2}O$) x (44 kJ/ 1mol $H_{2}O$)
= 6.8 kJ.
Hence the amount of heat required for vaporization of 2.8 g of $H_{2}O$ is 6.8 kJ.
