How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/g. oC over the entire liquid range, that the specific heat of steam is 1.99 J/g. oC, and the heat of vaporization of water is 40.79 kJ/mol

The amount of heat required is 775.7 kJ

Explanation:

The processes involved in the given problem are:  

Now, we calculate the amount of heat released or absorbed in all the processes.

where,

= amount of heat absorbed = ?

m = mass of water or ice = 248 g

= latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

where,

= amount of heat absorbed = ?

= specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

= final temperature =

= initial temperature =

Putting all the values in above equation, we get:

where,

= amount of heat absorbed = ?

m = mass of water or ice = 248 g

= latent heat of vaporization =      (Conversion factor used:  1 kJ = 1000 J and molar mass of water = 18 g/mol)

Putting all the values in above equation, we get:

where,

= amount of heat absorbed = ?

= specific heat of steam = 1.99 J/g°C

m = mass of water = 248 g

= final temperature =

= initial temperature =

Putting all the values in above equation, we get:

Calculating the total heat absorbed, we get:

Hence, the amount of heat required is 775.7 kJ