Question

The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from

50 °C to 5 °C, the surrounding 7.80 x 103 g of air absorbs the heat. The specific heat of
air is 1.01J/(9•°C). What is the change in air temperature, assuming all the heat released
by the ethanol is absorbed by the air?

Answer 1

a) heat it from 23.0 to 78.3

q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =

b) boil it at 78.3

(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =

c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ

Step-by-step explanation: