標題:
關於物理的簡單問題
1. A metal cup with heat capacity 200 J℃-1 contains hot coffee with mass 300 g. The initial temperature of the coffee is 80℃. When 100 g of milk at 20℃ is added into the coffee, what will be the final temperature of the mixture? The specific heat capacities of milk and coffee are respectively 4200 Jkg-1℃-1 and... 顯示更多 1. A metal cup with heat capacity 200 J℃-1 contains hot coffee with mass 300 g. The initial temperature of the coffee is 80℃. When 100 g of milk at 20℃ is added into the coffee, what will be the final temperature of the mixture? The specific heat capacities of milk and coffee are respectively 4200 Jkg-1℃-1 and 4300 Jkg-1℃-1. A. 62.7℃ B. 63.3℃ C. 66.3℃ D. 66.8℃ 2. A heater takes 2 minutes to heat up a liquid of mass 1.3 kg, from 30℃ to 40℃. If the heater takes further 24 minutes to completely boil off all the liquid, what is the specific latent heat of vaporization of that liquid? The boiling point and the specific heat capacity of the liquid are respectively 60℃ and 3200 Jkg-1℃-1. A. 3.2 x 105 Jkg-1 B. 3.84 x 105 Jkg-1 C. 4.16 x 105 Jkg-1 D. 4.99 x 105 Jkg-1 3. Which of the following can be the unit of energy? (1) kgm2s-2 (2) AVs (3) VC A. (1) and (2) only B. (1) and (3) only C. (2) and (3) only D. (1), (2) and (3) ※試列明計算步驟及各點對與錯之原因※
最佳解答:
1) Suppose there's no heat gain/loss to/from the surroundings, then Heat loss by coffee + cup = Heat gain by milk Suppose that final temp. = T ℃, then Heat loss by coffee = (200 + 0.3 x 4300)(80 - T) Heat gain by milk = 0.1 x 4200(T - 20) So, (200 + 0.3 x 4300)(80 - T) = 0.1 x 4200(T - 20) 1490(80 - T) = 420(T - 20) 119200 - 1490T = 420T - 8400 1910T = 127600 T = 66.8 2) Suppose there's no heat gain/loss to/from the surroundings, then Power of heater = 1.3 x 3200 x (40 - 30)/(2 x 60) = 346.7 W Also, using that heater, 2 minutes can raise the liquid's temp. by 10℃ and hence from 40℃ to 60℃, it takes 4 minutes. Thus for the remaining 20 minutes, all heat supplied by the heater were used to boil away the liquid, with energy = 346.7 x 20 x 60 = 416000 J Hence specific latent heat of vapourization = 416000/1.3 = 320000 J/kg 3) By relation kinetic energy = mv2/2, we can deduce that kg (m/s)2 = J So (1) correct. By electric power = Voltage x Current, we have Watt = VA So J = Ws = VAs So (2) is correct. By the definition that Voltage = Energy per Charge, we have V = J/C and hence J = VC So (3) is correct.
其他解答:8758B59A7FA1EEA7
關於物理的簡單問題
- 地球有多大-@1@
- 邊度有得買「御我」作既書---
- 渣打2888 換股價即係點？@1@
- 請問我海洋大學通訊 備取二十初 這樣有可能會上嗎 雖然去年前年都備到最後 但不知道今年變數會不會太大-
- maths(代數)@1@
- 請問香港那裡有比較浪漫的餐廳-
- -----F.1 MATHS-----10點@1@
- 請問夜間勤益進的去嗎-
- 玩Free Style會lag、魔獸爭霸也會
- 富比世公布2007年二千大企業,台灣哪42家企業進榜-
此文章來自奇摩知識+如有不便請留言告知
發問:1. A metal cup with heat capacity 200 J℃-1 contains hot coffee with mass 300 g. The initial temperature of the coffee is 80℃. When 100 g of milk at 20℃ is added into the coffee, what will be the final temperature of the mixture? The specific heat capacities of milk and coffee are respectively 4200 Jkg-1℃-1 and... 顯示更多 1. A metal cup with heat capacity 200 J℃-1 contains hot coffee with mass 300 g. The initial temperature of the coffee is 80℃. When 100 g of milk at 20℃ is added into the coffee, what will be the final temperature of the mixture? The specific heat capacities of milk and coffee are respectively 4200 Jkg-1℃-1 and 4300 Jkg-1℃-1. A. 62.7℃ B. 63.3℃ C. 66.3℃ D. 66.8℃ 2. A heater takes 2 minutes to heat up a liquid of mass 1.3 kg, from 30℃ to 40℃. If the heater takes further 24 minutes to completely boil off all the liquid, what is the specific latent heat of vaporization of that liquid? The boiling point and the specific heat capacity of the liquid are respectively 60℃ and 3200 Jkg-1℃-1. A. 3.2 x 105 Jkg-1 B. 3.84 x 105 Jkg-1 C. 4.16 x 105 Jkg-1 D. 4.99 x 105 Jkg-1 3. Which of the following can be the unit of energy? (1) kgm2s-2 (2) AVs (3) VC A. (1) and (2) only B. (1) and (3) only C. (2) and (3) only D. (1), (2) and (3) ※試列明計算步驟及各點對與錯之原因※
最佳解答:
1) Suppose there's no heat gain/loss to/from the surroundings, then Heat loss by coffee + cup = Heat gain by milk Suppose that final temp. = T ℃, then Heat loss by coffee = (200 + 0.3 x 4300)(80 - T) Heat gain by milk = 0.1 x 4200(T - 20) So, (200 + 0.3 x 4300)(80 - T) = 0.1 x 4200(T - 20) 1490(80 - T) = 420(T - 20) 119200 - 1490T = 420T - 8400 1910T = 127600 T = 66.8 2) Suppose there's no heat gain/loss to/from the surroundings, then Power of heater = 1.3 x 3200 x (40 - 30)/(2 x 60) = 346.7 W Also, using that heater, 2 minutes can raise the liquid's temp. by 10℃ and hence from 40℃ to 60℃, it takes 4 minutes. Thus for the remaining 20 minutes, all heat supplied by the heater were used to boil away the liquid, with energy = 346.7 x 20 x 60 = 416000 J Hence specific latent heat of vapourization = 416000/1.3 = 320000 J/kg 3) By relation kinetic energy = mv2/2, we can deduce that kg (m/s)2 = J So (1) correct. By electric power = Voltage x Current, we have Watt = VA So J = Ws = VAs So (2) is correct. By the definition that Voltage = Energy per Charge, we have V = J/C and hence J = VC So (3) is correct.
其他解答:8758B59A7FA1EEA7
文章標籤
全站熱搜
留言列表
