The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?

The correct answer would be - 2.4KJ or, 2400J

Explanation:

Given:

heat capacity of liquid water - 4.18 J/g·°C

heat of vaporization - 40.7 kJ/mol

Mass of water = 1g

Moles of water = mass/molar mass

= 1g/18.016g

= 0.055 moles

Then,

Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)

= m *s*dt + moles * heat of vaporization

= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ

= 137.94J + 2.26KJ

=0.138KJ + 2.26KJ

=2.4KJ or, 2400J

Thus, the correct answer would be - 2.4KJ or, 2400J

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.