close
標題:

物理(熱) 2題

發問:

1.) 要將250g水由80*C冷卻至0*C 要加入多少 -4*C的冰? (冰的比熱容量2060Jk-1*C-1) 2.) 把功率為1KW的電熱器浸在水中(0.5kg, 20*C) 然後開電熱器. 10min後有多少水變成蒸汽?

最佳解答:

 

此文章來自奇摩知識+如有不便請留言告知

1. Heat absorbed by water when it cools from 80'C to 0'C = (250/1000) x 4200 x 80 J = 84000 J where 4200 J/kg-C is the specific heat capacity of water Let m be the mass of ice needed, hence, heat released by ice = m x 2060 x (0 - (-4)) J = 8240 J Assume no heat lost to the surroundings, 84 000 = 8240m i.e. m = 10.19 kg 2. Heat produced by the heater in 10 minutes = 1000 x (10x60) J = 600 000 J Heat needed to raise the temperature of water from 20'C to 100'C = 0.5 x 4200 x (100 - 20) J = 168 000 J Hence, heat left to vapourize water to steam = (600 000 - 168 000) J = 432 000 J Mass of water that has turned into steam = 432000/2260000 kg = 0.19 kg where 2 260 000 J/.kg is the latent heat of vapourization of water

其他解答:

1) 設需要冰的質量為 m[解] 80*C 的水變成 0*C 的水放出能量 E1 = 0.250 x 4200 x (80 - 0) = 84000J冰會由 -4*C 變成 0*C 的冰,再變成 0*C 的水所放出的能量為(設冰的熔解比潛熱為 336 000 J / kg,需要能量 E2 = m x 2060 x (0 - (-4)) + m x 336000 = 344240m若過程中沒有能量的吸收或散失則吸熱 = 失熱 E1 = E2 84000 = 344240m m = 0.244kg(244g)(答)最小需要 244g 的冰****************************** 2) 設有 m kg 的水變成水蒸汽。[解] 10 分鐘電熱器放出的熱能 E1 = P t = 1000 x 10x60 = 600000J0.5kg(20*C)的水變成100*C 的水需要能量 E2 = 0.5 x 4200 x (100 - 20) = 168000Jm kg 100*C 的水變成 100*C 的水蒸汽需要能量(設水的汽化比潛熱為 2260000 J/kg) E3 = m x 2260000 = 2260000m若無能量的流失,則由電熱器發出的熱能 E1,會變被水吸收 E2,及使部份的水變成水蒸汽 E3,所以 E1 = E2 + E3 600000 = 168000 + 2260000m m = 0.191kg(答)有0.191kg 的水變成水蒸汽。960E6207E048F01D
arrow
arrow

    bxzbthf 發表在 痞客邦 留言(0) 人氣()