Answer
The amount of heat required for vaporization of 4.86 g of $H_{2}O$ is 11.9 kJ.
Work Step by Step
The amount of heat required to vaporize one mole of liquid is defined as heat of vaporization. Heat of vaporization is represented by dH vap.
dH vap of water is 40.7 kJ/mole.
Given the mass of sweat is 4.86 g, and it vaporizes at 25c.
So, 1 mol of water = 18.02 g of water and dH vap of water = 44.0 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 4.86 g of $H_{2}O$ can be calculated as follows:
= (4.86 g $H_{2}O$) x(1mol $H_{2}O$/18.02 g $H_{2}O$) x (44 kJ/ 1mol $H_{2}O$)
= 11.9 kJ.
Hence the amount of heat required for vaporization of 4.86 g of $H_{2}O$ is 11.9 kJ.
