Respuesta :
Answer: The heat released for the given process is -1892 kJ
Explanation:
The processes involved in the given problem are:
[tex]1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)[/tex]
Pressure is taken as constant.
To calculate the amount of heat released at same temperature, we use the equation:
[tex]q=m\times L_{f,v}[/tex] ......(1)
where,
q = amount of heat released = ?
m = mass of water/ice
[tex]L_{f,v}[/tex] = latent heat of fusion or vaporization
To calculate the amount of heat released 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 released = ?
[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
Calculating the heat absorbed for each process:
- For process 1:
Converting the latent heat of fusion in J/kg, we use the conversion factor:
1 kJ = 1000 J
So, [tex](\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg[/tex]
We are given:
[tex]m=2.6kg\\L_f=-334\times 10^3J/kg[/tex]
Putting values in equation 1, we get:
[tex]q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J[/tex]
- For process 2:
We are given:
[tex]m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K[/tex]
Putting values in equation 2, we get:
[tex]q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J[/tex]
Total heat absorbed = [tex]q_1+q_2[/tex]
Total heat absorbed = [tex][-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ[/tex]
Hence, the heat released for the given process is -1892 kJ
