Heat Energy Absorption Calculation for Water Boiling

How much heat energy must 150.0 g of water absorb to boil away completely?

Use the formula: mc001-1. jpg. 339.2 kJ 381.6 kJ 610.5 kJ 6,105 kJ.

Final answer:

To determine the heat energy absorbed by 150.0 g of water to completely boil away, we first convert grams to moles and then use the formula for heat absorption. In this case, 150.0 g of water must absorb 339.2 kJ of heat energy to completely boil.

Explanation:

In order to determine the amount of heat energy needed to completely boil away 150.0 g of water, we use the formula Q = mLv, where m is the mass of the substance, L is the heat of vaporization and Q is the heat absorbed. From the information given, we already know the heat of vaporization for water is 40.7 kJ/mol and we have 150.0 g of water.

First, we need to convert grams to moles using the molar mass of water, which is approximately 18 g/mol. This gives us 150.0 g / 18 g/mol = 8.33 mol.

Then, we can use Q = mLv to find the heat absorbed. So, Q = 8.33 mol x 40.7 kJ/mol = 339.2 kJ. Therefore, 150.0 g of water must absorb 339.2 kJ of heat energy to boil away completely.