How much heat is required to boil 98.5 g ?

Question

How much heat is required to boil 98.5 g of water at its boiling point?

Answer 1

To solve this you need the heat of vaporization of water. The heat of vaporization of water at its boiling point (100C) is 40.7 kj/mol.

Then use the heat of vaporization of water to calculate the amount of heat

98.5g H₂O x 1 mole H₂O x 40.7 kJ = 222.5 kJ or 222,500 J

18.02 g H₂O 1 mole H₂O

Answer 2

In this question, we need to use the formula: Q = MCAT At least that's the way I like to remember it having taken the MCAT myself.

In reality, the formula is Q = mC(delta T), with Q representing heart, m being mass in grams, C is the heat capacity of water, which is 4.186 Joules per gram per degree Celsius, and delta T is the change in temperature. You can also write this as T final - T initial.

We need to first fill out the variables we know. I already mentioned that C is equal to 4.186 Joules per gram per degree Celsius, we know that mass is 98.5 grams.

Finally, for temperature, we might find ourselves stuck, but we have to remember that room temperature is typically around 25˚ Celsius, which we will use for our initial temperature. And the boiling point of water is 100˚C, which is going to be the final temperature. Thus, we will have our delta T equal 100˚C - 25˚C - 75 as our delta T.

Now we just plug and chug.

Q = (98.5 g)(4.186 J/g˚C)(75)

Q = 30, 924.075 J or 30.92 kJ will be the amount of heat it takes for 98.5 g of water to reach boiling point from room temperature.