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


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> 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)
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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?


