Question

How much glucose (in g) must be metabolized to completely evaporate 1.00 g of water at 37 °C, given that ∆_cH_glucose = -2803.00 kJ mol^-1? In other words how much glucose is needed to evaporate water at 37 °C where the vapour is at 37 °C too.

For H₂O(l) at 100 °C ∆_vH = 40.67 kJ mol^-1; C_p(H₂O(l)) = 75.3 J mol^-1 K^-1; C_p(H₂O(g)) = 33.6 J mol^-1 K^-1.

(Previous post are wrong, please specify step by step)

		5.89g
		6.47g

I know this answer are wrong in case you get the same, and what confuses me is the fact that he's describing 3 processes 1) water 37-->100 2) water --> vapor 3) vapor water 100 --> 37. am I suppose to substract 3) from 1) and 2) ??

Answer 1