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標題:
phy .. heat
發問:
Simon is preparing a cup of ice tea at 0 C. He needs to put ice cubes at 0 C into 250 ml hot tea at 50 C. How many ice cubes should Simon put into the tea?Given: mass of 1 ml of tea = 1gmass of an ice cube = 40 gspecific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1specific heat capacity of water = 4200... 顯示更多 Simon is preparing a cup of ice tea at 0 C. He needs to put ice cubes at 0 C into 250 ml hot tea at 50 C. How many ice cubes should Simon put into the tea? Given: mass of 1 ml of tea = 1g mass of an ice cube = 40 g specific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1 specific heat capacity of water = 4200 J kg^-1 C^-1 specific heat capacity of tea = 4200 J kg^-1 C^-1 更新: Why no need to plus mc△T in the side of energy gained
最佳解答:
Let n be the number of ice cube (0.001 x 250)(4200)(50 - 0) = (0.04n)(3.34 x 10^5) n = 3.93 Simon should put 4 ice cube ito the tea 2007-01-10 03:09:08 補充： 如自再加埋個mc△T係energy gained 度咁個final溫度咪唔係0度囉又或者我咁計俾你睇energy lose = energy gainmc△T = ml + mc△T(0.001 x 250)(4200)(50 - 0) = (0.04n)(3.34 x 10^5) + (0.04n)(4200)(0 - 0)這個mc△T都係冇左n = 3.93答案都係一樣
phy .. heat
發問:
Simon is preparing a cup of ice tea at 0 C. He needs to put ice cubes at 0 C into 250 ml hot tea at 50 C. How many ice cubes should Simon put into the tea?Given: mass of 1 ml of tea = 1gmass of an ice cube = 40 gspecific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1specific heat capacity of water = 4200... 顯示更多 Simon is preparing a cup of ice tea at 0 C. He needs to put ice cubes at 0 C into 250 ml hot tea at 50 C. How many ice cubes should Simon put into the tea? Given: mass of 1 ml of tea = 1g mass of an ice cube = 40 g specific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1 specific heat capacity of water = 4200 J kg^-1 C^-1 specific heat capacity of tea = 4200 J kg^-1 C^-1 更新: Why no need to plus mc△T in the side of energy gained
最佳解答:
Let n be the number of ice cube (0.001 x 250)(4200)(50 - 0) = (0.04n)(3.34 x 10^5) n = 3.93 Simon should put 4 ice cube ito the tea 2007-01-10 03:09:08 補充： 如自再加埋個mc△T係energy gained 度咁個final溫度咪唔係0度囉又或者我咁計俾你睇energy lose = energy gainmc△T = ml + mc△T(0.001 x 250)(4200)(50 - 0) = (0.04n)(3.34 x 10^5) + (0.04n)(4200)(0 - 0)這個mc△T都係冇左n = 3.93答案都係一樣
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