Respuesta :
Answer: The heat required for the given process is 2659.3 kJ
Explanation:
The processes involved in the given problem are:
[tex]1.)H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\2.)H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\3.)H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\4.)H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\5.)H_2O(g)(100^oC)\rightarrow H_2O(g)(126^oC)[/tex]
Pressure is taken as constant.
To calculate the amount of heat absorbed at different temperature, we use the equation:
[tex]q=m\times C_{p,m}\times (T_{2}-T_{1})[/tex] .......(1)
where,
q = amount of heat absorbed = ?
[tex]C_{p,m}[/tex] = specific heat capacity of medium
m = mass of water/ice
[tex]T_2[/tex] = final temperature
[tex]T_1[/tex] = initial temperature
To calculate the amount of heat released at same temperature, we use the equation:
[tex]q=m\times L_{f,v}[/tex] ......(2)
where,
q = amount of heat absorbed = ?
m = mass of water/ice
[tex]L_{f,v}[/tex] = latent heat of fusion or vaporization
Calculating the heat absorbed for each process:
- For process 1:
We are given:
[tex]m=856g\\C_{p,s}=2.03J/g^oC\\T_1=-18^oC\\T_2=0^oC[/tex]
Putting values in equation 1, we get:
[tex]q_1=856\times 2.03J/g^oC\times (0-(-18))^oC\\\\q_1=31278.24J[/tex]
- For process 2:
Converting the latent heat of fusion in J/g, we use the conversion factor:
Molar mass of water = 18 g/mol
1 kJ = 1000 J
So, [tex](\frac{6.01kJ}{1mol})\times (\frac{1000J}{1kg})\times (\frac{1mol}{18g})=334J/g[/tex]
We are given:
[tex]m=856g\\L_f=334J/g[/tex]
Putting values in equation 2, we get:
[tex]q_2=856g\times 334J/g=285904J[/tex]
- For process 3:
We are given:
[tex]m=856g\\C_{p,l}=4.184J/g^oC\\T_1=0^oC\\T_2=100^oC[/tex]
Putting values in equation 1, we get:
[tex]q_3=856g\times 4.184J/g^oC\times (100-(0))^oC\\\\q_3=358150.4J[/tex]
- For process 4:
Converting the latent heat of vaporization in J/g, we use the conversion factor:
Molar mass of water = 18 g/mol
1 kJ = 1000 J
So, [tex](\frac{40.79kJ}{1mol})\times (\frac{1000J}{1kg})\times (\frac{1mol}{18g})=2266J/g[/tex]
We are given:
[tex]m=856g\\L_v=2266J/g[/tex]
Putting values in equation 2, we get:
[tex]q_4=856g\times 2266J/g=1939696J[/tex]
- For process 5:
We are given:
[tex]m=856g\\C_{p,g}=1.99J/g^oC\\T_1=100^oC\\T_2=126^oC[/tex]
Putting values in equation 1, we get:
[tex]q_5=856g\times 1.99J/g^oC\times (126-(100))^oC\\\\q_5=44289.44J[/tex]
Total heat absorbed = [tex]q_1+q_2+q_3+q_4+q_5[/tex]
Total heat absorbed = [tex][31278.24+285904+358150.4+1939696+44289.44]J=2659318.08J=2659.3kJ[/tex]
Hence, the heat required for the given process is 2659.3 kJ
