物理(熱) 2題

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

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。