A food product is being frozen in a system capable of removing 6000 kJ of thermal energy. The product has a specifi c heat of 4 kJ/(kg C) above the freezing temperature of 2 C, the latent heat of fusion equals 275 kJ/kg, and the frozen product has a specifi c heat of 2.5 kJ/(kg C) below 2 C. If 10 kg of product enters the system at 10C, determine the exit temperature of the product..

Respuesta :

Answer : The exit temperature of the product is, [tex]-35.2^oC[/tex]

Explanation :

Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen

[tex]Q=m\times c_1\times (T_2-T_1)+m\times L_f+m\times c_2\times (T_4-T_3)[/tex]

where,

Q =  Total heat = 6000 kJ

m = mass of product = 15 kg

[tex]c_1[/tex] = specific heat of liquid = [tex]4kJ/kg^oC[/tex]

[tex]L_f[/tex] = latent heat of fusion = [tex]275kJ/kg[/tex]

[tex]c_2[/tex] = specific heat of frozen = [tex]2.5kJ/kg^oC[/tex]

[tex]T_1[/tex] = initial temperature of liquid = [tex]2^oC[/tex]

[tex]T_2[/tex] = final temperature of liquid = [tex]10^oC[/tex]

[tex]T_3[/tex] = initial temperature of frozen = ?

[tex]T_4[/tex] = final temperature of frozen = [tex]2^oC[/tex]

Now put all the given value in the above expression, we get:

[tex]6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC][/tex]

[tex]T_3=-35.2^oC[/tex]

Thus, the exit temperature of the product is, [tex]-35.2^oC[/tex]

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