中三phy問題

解：混合物溫度
m(1)c(1)ΔT=mL+m(2)c(2)ΔT
0.5 x 4200 x(70-T) =0.2 x 3.34 x 10^5 + 0.2 x 4200 x (T-0)
147000-2100T=66800+840T
80200=2940T
T=27.3℃


混合物的公式：m(1)c(1)ΔT=m(2)c(2)ΔT
mL：溶解比潛熱

設混合後的溫度是t :
用到的公式:E=mc*T , E=ml
吸熱=散熱
(70-t)*0.5*4200=0.2*334000+4200*0.2*t
147000-2100t=66800+840t
147000-66800=840t+2100t
80200=2940t
t=27.27891...
所以混合後的溫度是27.3度(三位有效數字)

Assume the specific heat capacity of water = 4200 J/kg ℃
Assume the specific latent heat of fusion of ice = 3.34 x 10^5 J/kg
Assume no energy is lost to the surrounding.

Let the final temperature of the mixture be T

Energy absorbed by the 0℃ melting ice:
=specific latent heat of fusion of ice
=(0.2)(3.34 x 10^5)
=66800 J

Energy absobed by the 0℃ water:
=specific heat capacity of water
=(0.2)(4200)(T)

Energy released by the 70℃ water:
=specific heat capacity of water
=(0.5)(4200)(70-T)

By the law of conservation of energy,

heat released = heat absorbed

(0.5)(4200)(70-T) = 66800 + (0.2)(4200)(T)

T=19.10℃

∴The final temperature of the mixture is 19.10℃

2007-04-02 00:29:07 補充：
Sorry, I wrongly take the 0.2 as 0.5T should be equals to 27.28℃Others remain unchanged.

由於一面吸入既能量 = 一面放出既能量
設混合後的溫度為T
(0.5)(4200)(70-T) = (0.2)(4200)(T-0) + (0.2)(3.34x10^5)
運算中.......
T = 27.28度C