Question

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

Answer 1

Answer: The amount of heat required is 775.7 kJ

Step-by-step explanation:

The processes involved in the given problem are:

`1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)`

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

• For process 1:

`q_1=m* L_f`

where,

`q_1` = amount of heat absorbed = ?

m = mass of water or ice = 248 g

`L_f` = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

`q_1=248g* 334J/g=84832J`

• For process 2:

`q_2=m* C_(p,l)* (T_(2)-T_(1))`

where,

`q_2` = amount of heat absorbed = ?

`C_(p,l)` = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

`T_2` = final temperature =
`100^oC`

`T_1` = initial temperature =
`0^oC`

Putting all the values in above equation, we get:

`q_2=248g* 4.184J/g^oC* (100-0)^oC=103763.2J`

• For process 3:

`q_3=m* L_v`

where,

`q_3` = amount of heat absorbed = ?

m = mass of water or ice = 248 g

`L_v` = latent heat of vaporization =
`40.79kJ/mol* (1000)/(18)=2266.1J/g` (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)

Putting all the values in above equation, we get:

`q_3=248g* 2260J/g=560480J`

• For process 4:

`q_4=m* C_(p,g)* (T_(2)-T_(1))`

where,

`q_4` = amount of heat absorbed = ?

`C_(p,g)` = specific heat of steam = 1.99 J/g°C

m = mass of water = 248 g

`T_2` = final temperature =
`154^oC`

`T_1` = initial temperature =
`100^oC`

Putting all the values in above equation, we get:

`q_4=248g* 1.99J/g^oC* (154-100)^oC=26650.1J`

Calculating the total heat absorbed, we get:

`Q=q_1+q_2+q_3+q_4`

`Q=[84832+103763.2+560480+26650.1]J=775,725.3J=775.7kJ`

Hence, the amount of heat required is 775.7 kJ