How much heat does your body lose when 3.15g of sweat evaporates from your skin at 25 ∘C? (Assume that the sweat is only water.) The heat of vaporization of water at 25 ∘C is 44.0 kJ/mole?
Please someone help explain how to set up this problems or how to do it
It's the old q = m*heat vaporization
m = 3.15g/18 = ? mols H2O
heat vap = 44.0 kJ/mol
q = ? kJ.
To solve this problem, we can use the equation:
q = n * ΔH
where q is the amount of heat, n is the number of moles, and ΔH is the heat of vaporization.
First, we need to calculate the number of moles of water that evaporated. To do this, we can use the molar mass of water, which is approximately 18 g/mol:
n = mass / molar mass = 3.15 g / 18 g/mol
Next, we can convert the moles of water to kilojoules of heat using the given heat of vaporization:
q = n * ΔH = (3.15 g / 18 g/mol) * 44.0 kJ/mol
Now, let's perform the calculations step by step:
1. Calculate the number of moles:
n = 3.15 g / 18 g/mol ≈ 0.175 mol
2. Calculate the heat:
q = 0.175 mol * 44.0 kJ/mol ≈ 7.7 kJ
Therefore, your body loses approximately 7.7 kJ of heat when 3.15 g of sweat evaporates from your skin at 25 ∘C.