Question

You are ready to contrast the efficiency of a new technology against good ‘ol water boiling. To do so you first must calculate values for that process. Given a closed system (like a pot with a lid) of 1kg of liquid water at 100C,

What is the work done to convert all the water to steam?

How much heat is transferred into the system to make this happen? (Recall, look

up constants as needed)

What is the change in internal E?

Answer part b again if the pot started at room temperature.

Answer 1

Volume of liquid water at 100 deg.c =1.044 cm3/g and Volume of steam = 1673cm3/g

Change in volume = 1673-1.044 cm3=1671.956cm³/g

18 gm = 1 gmole

1gm= 1/18 moles

So change in volume = 1671.956*18cm3/gmoles=30095 cm3/gmoles

1000 gm of steam correespond to 1000/18= 55.5 gmoles

So volume change in total= 55.5*30095=1.67*106 cc=1.67*1000 L

Work done = P*change in volume = 1atm*1.67*1000L= 1670L.atm= 1670*101.3 Joules=169171 joules

change in internal energy ( from steam tables)= 2063 Kj/kg

Hence for 1 kg. change in internal energu= 2063*1000 Joules

deltaU= Q+W ( from 1st law of thermodynamics)

2063*1000= Q+169171

Q= 2063*1000-169171= 1893829 joules

2. When temperature is 25 deg.c, change in volume = (1673-1)*18cm³*1000 /1000 L = 1672 L

work done = 1atm*1672 L = 1672L.atm= 1672*101.3 Joules =169374 joules

change in internal energy = (2506.5-104.8)Kj/Kg*1 kg = 2400.8 Kj

Q= 2400.8*1000-169374 =2231426 Joules