Question

A 50.0 mL sample of water is heated to its boiling point. How much heat is required to vaporize it? (Assume a density of 1.00 g/mL.)

Express the heat in kilojoules to three significant figures.

a) 5.00 kJ

b) 50.0 kJ

c) 10.0 kJ

d) 4.18 kJ

Answer 1

The correct amount of heat required to vaporize a 50.0 mL sample of water is approximately 113 kJ, which is calculated by multiplying the mass of the water by the heat of vaporization. The mass of the water is 50.0 g, and the heat of vaporization is 2.26 × 10³ J/g.

Step-by-step explanation:

To calculate the amount of heat required to vaporize a 50.0 mL sample of water, we first need to determine the mass of the water, which we can find using the density of water (1.00 g/mL). The mass of the water is therefore 50.0 mL × 1.00 g/mL = 50.0 g. Next, we use the fact that the heat of vaporization of water is 2.26 × 10³ J/g. Therefore, we must multiply the mass of the water by this value to find the total heat required for vaporization:

Q = m × Lv = 50.0 g × 2.26 × 10³ J/g

Q = (50.0 × 2.26 × 10³) J = 113,000 J or 113 kJ. However, since the options provided are in somewhat smaller magnitudes, let's convert this to kilojoules:

Q = 113 kJ → 113 kJ (expressed in kilojoules to three significant figures).

The correct answer choice is not listed; however, based on our calculation, the correct amount of heat that is required to vaporize the 50.0 mL sample of water is approximately 113 kJ.