Question

A food product is being frozen in a system capable of removing 6000 kJ of thermal energy. The product has a specific 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 specific heat of 2.5 kJ/(kg °C) below - 2°C. If 10 kg of product enters the system at 10°C, determine the exit temperature of the product.

Answer 1

Answer:

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.

Answer 2

A food product is being frozen in a system

Find-We have to determine the exit temperature of the product

Explanation:

Given:-

Total Heat Q is=6000 kJ

Specific Heat of liquid $C_{1}=4 kJ/kg$

Latent Heat of Fusion $L_{f}=275 kJ/kg$

mass of the product m=15 kg

specific heat of frozen $C_{2}=2.5 kJ/kg'C$

Initial temperature of liquid $T_{1}$=-2° C

Final temperature of liquid $T_{2}$=10° C

Final Temperature of Frozen $T_{2}$=-2° C

For finding Temperature $T_{3}$ we use the formula of Total Heat

The Total Heat=Heat lost in the liquid + Latent Heat of fusion+Heat lost by fusion

$Q=m\times C_{1} \times(T_{2}-T_{1} )+m\times L_{f}+m \times C_{2} \times(T_{4}-T_{3} )$

$6000 kJ=[15 kg\times4kJ/kg\times(10-(-2)]+[15kg\times275kJ/kg]+[15 kg\times2.5\times(-2-T_{3} )]\\6000 kJ=4845+37.5(-2-T_{3}) ]\\1155=37.5(-2-T_{3} )\\T_{3}=-32.8 'C$

Hence, the exit temperature of the product=-32.8° C