What mass of water would need to evaporate from your skin in order to dissipate 1.70 ×10⁵ J of heat from the surface of your body? H₂O(l) → H₂O(g) ΔH_vap = 40.7 kJ/mol A) 2.26 g B) 4.18 g C) 75.2 g D) 4.18 ×10³g E) 4.07 × 10⁴ g

Question

Quizplus · Accepted Answer

First, we need to convert the heat dissipated from the body from joules to kilojoules:
1.70 ×10^15 J ÷ 1000 = 1.70 ×10^12 kJ

Next, we can use the given molar heat of vaporization of water to calculate the amount of water that needs to evaporate:
40.7 kJ/mol × x mol = 1.70 ×10^12 kJ
x = 4.18 × 10^10 mol

Finally, we can convert moles of water to grams of water using the molar mass of water:
4.18 × 10^10 mol × 18.015 g/mol = 7.52 × 10^11 g
or
75.2 g

Therefore, the mass of water that needs to evaporate from the skin is approximately 75.2 g. Option C is the correct choice.

What Mass of Water Would Need to Evaporate from Your