# Calculate the energy in the form of heat (in kJ) required to convert 325 grams of liquid water at 20.0 °C to steam at 115 °C

Calculate the energy in the form of heat (in kJ) required to convert 325 g of liquid water at 20.0 °C to steam at 115 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: liquid water = 4.184 J/(g K), steam = 1.92 J/(g K)

Here is what I did:

$$Q=mcT$$

\begin{align}Q &= 325 ( 4.184) (100 - 20)\\ &= 325 ( 4.184 ) ( 80)\\ &= 108784\ \mathrm J\end{align}

\begin{align}Q&= ml\\ Q&= 325 ( 2256)\\ &= 733200\ \mathrm J\end{align}

\begin{align}Q &= 325 (1.92) ( 115 - 100)\\ &=325 (1.92) ( 15)\\ &= 9360\ \mathrm J\end{align}

$$(108784+733200+9360) \mathrm{J}= 851\ \mathrm{kJ}$$

But the correct answer is $$735\ \mathrm{kJ}$$

Do you know the right way to do it?

• Always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation. – Loong Apr 30 at 17:15
• If you mess up the first part by a factor 10 and switch two digits when writing down the answer, you get 735 kJ. I would not call that correct. – Karsten Theis Apr 30 at 17:28