Answer to Question #162482 in General Chemistry for Rachel

Question

1. Steam at 100^oC was passed into a flask containing 275 grams of water at 21 ^oC where the steam condensed. How many grams of steam must have condensed if the temperature of the water in the flask was raised to 83 ^oC? The heat of vaporization of water at 100 ^oC is 40.7 kj/mole and the specific heat constant is 4.18 Jules/(gram ^oC)

Answer 1

Q = mass× specific heat × ( final temp - initial temp )

Q = 275g × {4.18 Jules/(gram⁰C) } × {(83 - 21 ) ⁰C }

Q = 71269 J

Q = 71.269 KJ

heat released (Q) = (40.7 KJ/mol )× no. of moles (n)

71.269 KJ = 40.7 KJ/mol × n

n = 1.75 mol

now no. of mol × mass of water

= 1.75×18

= 31.5 g