ZJSG09112007
Answered

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What mass of ice at -14°C will be needed to cool 200cm^3 of an orange drink(essentially water) from 25°C. to 10°C.
(Specific Latent heat of Fusion of ice= 3.36 ×10^5 J/Kg
Specific heat capacity of ice =2100 J/kg/k
specific heat capacity of water=4200J/Kg/k) ?

Please show workings.

Sagot :

sqdancefan

Answer:

  31 grams

Explanation:

The amount of heat required to raise a mass (m, in kg) of ice from -14 °C to 10 °C is ...

  heat to raise temperature of ice to 0 °C:

     (2100 J/kg/K)(m kg)(14 K) = (29400m) J

  heat to melt ice:

     (336000 J/kg)(m kg) = (336000m) J

  heat to raise melted ice to 10 °C

     (4200 J/kg/K)(m kg)(10 K) = (42000m) J

So, the total heat required to raise m kg of ice from -14 °C to +10 °C is ...

  (29,400 + 336,000 + 42,000)m J = (407400m) J

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The heat required to cool 200 cc = 200 g = 0.2 kg of drink by 15 K is ...

  (4200 J/kg/K)(0.20 kg)(15 K) = 12600J

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The minimum amount of ice will be used if it all melts, so the two quantities of heat are the same:

  407400m = 12600 . . . . joules

  m = 12600/407400 ≈ 0.030928 . . . . kg

About 31 grams of ice are needed.

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