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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )