Question

mic or endothermic 612. Ctus is expected. Use equations 6.4 to explain why. 3 ex6ther- farmers sometimes spray water on the fruit trees Citrus farmers frost is when a frost is e that the AvapH of an evaporating liquid comes 6.13. Assume that t the liquid itself, which then changes temperature ac- cording to equation AvapH-q mcAT lal Explain the presence of the negative sign in this equation. (b) If 1.00 g of H2O evaporates at 25.0°C in an adiabatic con- tainer initially containing 100.0 g of H20, what is the final temperature of the water? Use c 4.18 J/g°C and Δνα,H 43.99 kJ/mol at 25.0°C. The vapor pressure of H,0 at 25.0°C 5 23.77 torr, assume it remains constant throughout the tem perature change of the liquid. 6.14. As a made to freeze i ow-up to the previous exercise, liquids can be eze if enough vanor is forcibly removed from then

Answer 1

(a) The negative sign indicates that the process is exothermic and heat is released.

(b) As 1.00 g evaporates, it absorbs heat. Rest 99 g of water releases heat.

The temperature of 1.00 g water increases from 25.0 ⁰C to 100.0 ⁰ C. Heat required for the process = 1.00 g * 4.18 J/ g ⁰C * ( 100 - 25 ) ⁰C = 313.5 J

Then at 100 0C, 1.00 g water evaporates. Number of moles of water evaporates = 1.00 g / (18 g / mole) = 0.056 mole

Heat required for evaporation = 0.056 mole * 43.99 kJ/ mole = 2.46 kJ = 2460 J

Total heat required for the process = 313.5 + 2460 J = 2773.5 J

Heat lost by 99 g of water = 2773.5 J

So, - 2773.5 = 99 g * 4.18 J/ g ⁰C * ( Final Temperature - Initial temperature)

So, - 6.70 = Final Temperature - 25.0

So, Final temperature = 18.3 ⁰ C