標題:
S.3 Phy Latent Heat
發問:
(1) 1kg steam at 100℃ is mixed with 1 kg ice at 0℃, assumed there is no heat loses to surroundings, what will be the final temperature. In what state will be the result outcome? (Take specific heat capacity as 4200 Jkg-1℃-1, take latent of vaporization as 2.26 X 106 J kg-1 and fusion as 3.34 X 105 J kg-1)... 顯示更多 (1) 1kg steam at 100℃ is mixed with 1 kg ice at 0℃, assumed there is no heat loses to surroundings, what will be the final temperature. In what state will be the result outcome? (Take specific heat capacity as 4200 Jkg-1℃-1, take latent of vaporization as 2.26 X 106 J kg-1 and fusion as 3.34 X 105 J kg-1) (2) After some steam at 100 c is passes through blocks of ice at 0 c, 40 g of water is collected. Find the mass of the steam introduced.
最佳解答:
(1) Total energy released by steam to become all water = 2.26 * 10^6 * 1 = 2.26 * 10^6 J Total energy required to melt all ice = 3.34 * 10^5 * 1 = 3.34 * 10^5 J Let m be the mass of steam required to melt the ice. 2.26 * 10^6 * m = 3.34 * 10^5 m = 0.147788 So, all ice will melt into water. The final outcome will be a mixture of steam and water. Total energy released by remaining steam to become all water = 2.26 * 10^6 * (1-m) = 1925999.12 J Total energy required to raised the (m+1)kg water from 0 to 100 degree Celsius = 1.147788 * 4200 * 100 = 482070.96 Assume the final mixture will be 100 degree Celsius of steam and water. Let m2 be the total mass of steam which has transformed to water. 2.26 * 10^6 * m2 = 3.34 * 10^5 + (1+m2) * 4200 * 100 m=0.40978 kg So. the final mixture will be steam and water at 100 degree Celsius. (2) Let m be the mass of the steam introduced. 2.26 * 10^6 * m = 40 * 10^-3 * 3.34 * 10^5 m = 0.0059115 kg
其他解答:
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(1) Heat absorbed when 1 kg of ice completely melts then changes to 1kg of 100oC water = mLf + mcDT = (1) x (3.34 x 105) + (1) x (4200) x (100 - 0) = (3.34 x 105) + (4.2 x 105) = 7.54 x 105 J Heat released when 1 kg of steam completely changes to 100oC water = mLv = (1) x (2.26 x 106) = 2.26 x 106 J > 7.54 x 105 J Therefore, the steam would NOT completely condense, and the mixture contains 100oC steam and 100oC water. The final temperature = 100oC (2) Let the mass of steam introduced be y kg. Then, the mass of melted ice = (0.04 - y) kg. Heat released when y kg of steam condenses = mLv = (y) x (2.26 x 106) J Heat absorbed when (0.04 - y) kg of ice melts = (0.04 - y) x (3.34 x 106) J (y) x (2.26 x 106) = (0.04 - y) x (3.34 x 105) (2.26 x 106)y = (1.336 104) - (3.34 x 105)y (2.594 x 106)y = (1.336 104) y = 0.00515 Mass of steam = 0.00515 kg = 5.15 g =
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