close
標題:
關於heat既問題 (冰溝cola) 急!!
發問:
最佳解答:
first, we have to know wt is the mixture, is that all ice, ice and cola mixture or all cola. Then we, can guess, supose, it is a all ice(coz, u have 5kg of ice but just 0.3kg of cola) Applying energy conservation principal, energy gain by the ice=energy lost by the cola, let the final temperature be T, Energy gained by the ice=3.34*10^5 Jkg^-1 x 5=1670000J Energy lost by cola = 5300Jkg^-1 x 0.3 x (25-0)= 397500J So, it proves that my suggestion is correct, as you have not given the specific latent heat of cola, i can't calculate whether the mixture is all ice or ice-liquid, but nevermind, the T=0 and T won't change in reality, as the energy is not enough to melt the ice
其他解答:
m1 x c1 x ( T - T1 ) + m1 x l = m2 x c2 x ( T2 - T ) 5 x 4200 x T + 5 x 3.34 x 10^5 = 0.3 x 5300 x ( 25 - T ) 21000T + 1670000 = 39750 - 1590T T = -72.167oC (the ice have not melt) it is lower than reality
關於heat既問題 (冰溝cola) 急!!
發問:
此文章來自奇摩知識+如有不便請留言告知
Some ice cubes(5kg,0 degree celcius ) are added to a glass of coke(0.3kg 25degree celcius). If no heat is lost to the surrounding, what is the final temperature T of the mixture ? Is T higher or lower in reality?given that:specific latent heat of ice is 3.34*10^5 Jkg^-1specific heat capacity of water... 顯示更多 Some ice cubes(5kg,0 degree celcius ) are added to a glass of coke(0.3kg 25 degree celcius). If no heat is lost to the surrounding, what is the final temperature T of the mixture ? Is T higher or lower in reality? given that: specific latent heat of ice is 3.34*10^5 Jkg^-1 specific heat capacity of water is 4200Jkg^-1(degree celcius^-1) specific heat capacity of coke is 5300Jkg^-1(degree celcius^-1) thx 更新: 唔該 咁有冇辦法計到幾多冰會溶 幾多會唔溶?最佳解答:
first, we have to know wt is the mixture, is that all ice, ice and cola mixture or all cola. Then we, can guess, supose, it is a all ice(coz, u have 5kg of ice but just 0.3kg of cola) Applying energy conservation principal, energy gain by the ice=energy lost by the cola, let the final temperature be T, Energy gained by the ice=3.34*10^5 Jkg^-1 x 5=1670000J Energy lost by cola = 5300Jkg^-1 x 0.3 x (25-0)= 397500J So, it proves that my suggestion is correct, as you have not given the specific latent heat of cola, i can't calculate whether the mixture is all ice or ice-liquid, but nevermind, the T=0 and T won't change in reality, as the energy is not enough to melt the ice
其他解答:
m1 x c1 x ( T - T1 ) + m1 x l = m2 x c2 x ( T2 - T ) 5 x 4200 x T + 5 x 3.34 x 10^5 = 0.3 x 5300 x ( 25 - T ) 21000T + 1670000 = 39750 - 1590T T = -72.167oC (the ice have not melt) it is lower than reality
文章標籤
全站熱搜
留言列表
