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

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:

$q_1=m\times L_f$

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:

$q_1=248g\times 334J/g=84832J$

• For process 2:

$q_2=m\times C_{p,l}\times (T_{2}-T_{1})$

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:

$q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J$

• For process 3:

$q_3=m\times L_v$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_v$ = latent heat of vaporization = $40.79kJ/mol\times \frac{1000}{18}=2266.1J/g$      (Conversion factor used:  1 kJ = 1000 J and molar mass of water = 18 g/mol)

Putting all the values in above equation, we get:

$q_3=248g\times 2260J/g=560480J$

• For process 4:

$q_4=m\times C_{p,g}\times (T_{2}-T_{1})$

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:

$q_4=248g\times 1.99J/g^oC\times (154-100)^oC=26650.1J$

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