Respuesta :
Answer: The amount of heat required is 775.7 kJ
Explanation:
The processes involved in the given problem are:
[tex]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)[/tex]
Now, we calculate the amount of heat released or absorbed in all the processes.
- For process 1:
[tex]q_1=m\times L_f[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
m = mass of water or ice = 248 g
[tex]L_f[/tex] = latent heat of fusion = 334 J/g
Putting all the values in above equation, we get:
[tex]q_1=248g\times 334J/g=84832J[/tex]
- For process 2:
[tex]q_2=m\times C_{p,l}\times (T_{2}-T_{1})[/tex]
where,
[tex]q_2[/tex] = amount of heat absorbed = ?
[tex]C_{p,l}[/tex] = specific heat of water = 4.184 J/g°C
m = mass of water = 248 g
[tex]T_2[/tex] = final temperature = [tex]100^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]0^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J[/tex]
- For process 3:
[tex]q_3=m\times L_v[/tex]
where,
[tex]q_3[/tex] = amount of heat absorbed = ?
m = mass of water or ice = 248 g
[tex]L_v[/tex] = latent heat of vaporization = [tex]40.79kJ/mol\times \frac{1000}{18}=2266.1J/g[/tex] (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)
Putting all the values in above equation, we get:
[tex]q_3=248g\times 2260J/g=560480J[/tex]
- For process 4:
[tex]q_4=m\times C_{p,g}\times (T_{2}-T_{1})[/tex]
where,
[tex]q_4[/tex] = amount of heat absorbed = ?
[tex]C_{p,g}[/tex] = specific heat of steam = 1.99 J/g°C
m = mass of water = 248 g
[tex]T_2[/tex] = final temperature = [tex]154^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]100^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_4=248g\times 1.99J/g^oC\times (154-100)^oC=26650.1J[/tex]
Calculating the total heat absorbed, we get:
[tex]Q=q_1+q_2+q_3+q_4[/tex]
[tex]Q=[84832+103763.2+560480+26650.1]J=775,725.3J=775.7kJ[/tex]
Hence, the amount of heat required is 775.7 kJ
