¿Qué cantidad de calor debe extraer un refrigerador de una masa de 10 [kg] de agua inicialmente a 20 [ºC] para tener hielo a - 5 [ºC]? (calor latente del hielo es 80 [Kcal/kg ]), el calor específico del agua es 1 [Kcal /kg ºC] y del hielo es 0.5 [Kcal /kg ºC] )

Total heat extract = 1025 kCal

Step-by-step explanation

Ans

 

Mass of water M = 10 Kg

 

Initial temperature of water T_i = 20 ^oC

Final temperature T_f = - 5 ^oC

 

There are three processes

1

Water from 20 ^oC to 0 ^oC water

 

2

0 ^oC water to 0 ^oC ice

 

3

0 ^oC ice to - 5 ^oC ice

 

 

Now for Step 1

Heat extract

Q₁ = M ?×? C_w?×? (T_i - 0)

 

Where C_w = Specific heat of water = 1 kCal/Kg.^oC

 

Q₁ = 10 ?×? 1 ?×? (20 - 0)

 

Q₁ = 200 kCal

 

For Step 2

Heat extract

Q₂ = M ?×? L

 

Where L = latent heat of ice = 80 kCal/Kg

 

Q₂ = 10 ?×? 80 = 800 kCal

 

For Step 3

Heat extract

Q₃ = M ?×? C_ice?×? (0 - T_f)

 

Where C_ice = Specific heat of ice = 0.5 kCal/Kg.^oC

 

Q₃ = 10 ?×? 0.5 ?×? (0 - (- 5))

 

Q₃ = 10 ?×? 0.5 ?×? 5

 

Q₃ = 25 kCal

 

 

 

Total heat extract

Q = Q₁ + Q₂ + Q₃

 

Q = 200 + 800 + 25

 

Q = 1025 kCal

 

 

 

 

 

 

 

 

 

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