Question

8. What mass of water would need to evaporate from your skin inorder to dissipate 1.7 ´ 10⁵ J of heat from yourbody? (H₂O(l) ® H₂O(g)DH_vap = 40.7 kJ/mol; H₂O = 18.01g/mol)

A.     7.52 x 10⁴ g

B.     418 g

C.     75.3 g

D.     58.4 g

E.     6.92 x 10⁶ g

Answer 1

Given data

ΔHvap = 40.7 kJ/mol

              =40700 J/mol

The energy required = 1.7*10⁵ J

∴ Number of moles of water must beevoporated = 1.7*10⁵ J /  40700 J/mol

                                                                               =4.177 mol

The mass of water evoporated = moles * molar mass

                                                = 4.177 mol * 18.01 g/mol

                                                = 75.23 g

Hence option C is correct.