phy...

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2007-01-07 9:38 pm

The initial temperature of a jar of juice is 80 C and the mass of juice is 2 kg. Susan adds ice cubes into the jar in order to cool down the juice to 20 C . What is the minimum number of ice cubes at 0 C required?
(( neglect the heat capacity of the jar and assume there is no heat exchange with the surroundings))

更新1:

Given : Mass of each ice cube = 0.15 kg Specific heat capacity of juice = 4700 J kg^-1 C^-1 Specific heat capacity of water = 4200 J kg^-1 C^-1 Specific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1 p.s. C means degree celsius please solve this question...thz!!!

回答 (2)

Audrey Hepburn

2007-01-07 9:55 pm

✔ 最佳答案

Energy required to change 80°C juice to 20°C
= mc△T
= (2)(4700)(80 - 20)
= 564 000 Jkg-1°C-1
By the law of conservation of energy
Heat gain by the ice = heat loss by the juice
ml + mc△T = 564 000
m(l + c△T) = 564 000
m[3.34 X 10^5 + 4200(20 - 0)] = 564 000
m = 1.3493 kg
So, the minimum mass of ice cubes used is 1.3493 kg.
So, the minimum number of ice required
= 1.3493 / 0.15
= 8.99
= 9

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Ming

2007-01-07 9:55 pm

Energy required cool down the juice to 20 C
=2x4700x(80-20)
=564000

Let n be the number of ice cube,
Energy required the water 0 C to 20 C
=nx0.15x4200x20
=12600n

Energy required the ice cube 0 C to the water 0 C
=nx0.15x3.34x10^5
=50100n

12600n+50100n=564000
62700n=564000
n=8.99
therefore,n=9

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