International Baccalaureate · IB Diploma · Biology · SL Topik 1 4:05 Narasi bahasa Inggris · Subtitle bahasa Inggris + 中文 disematkan langsung
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A water droplet holds together on a leaf.
水滴在叶片上保持完整。
Its behaviour comes from interactions between molecules, not a skin made of solid water.
它的行为源于分子间的相互作用，而非由固态水构成的表皮。
Water molecules are polar.
水分子是极性的。
Attraction between the partially positive hydrogen of one molecule and a partially negative oxygen of another gives hydrogen bonding.
一个分子中部分带正电的氢与另一个分子中部分带负电的氧之间的吸引力形成了氢键。
Cohesion is attraction between molecules of the same substance, so water molecules hold on to each other.
内聚力是同种物质分子之间的吸引力，所以水分子彼此相连。
Adhesion is attraction between different substances, so water clings to a surface such as glass or a leaf.
黏附力是不同物质之间的吸引力，所以水会附着在玻璃或叶片等表面上。
The same intermolecular interactions help explain surface tension and the energy needed to change temperature.
相同的分子间作用力有助于解释表面张力以及改变温度所需的能量。
Heating changes molecular motion and interactions; it does not normally break the covalent O–H bonds.
加热会改变分子运动和相互作用；通常不会破坏共价O–H键。
In the lab: Compare equal drops on clean surfaces under fixed temperature.
在实验室中，比较相同液滴在洁净表面上的固定温度条件。
Measure contact angle or spreading consistently.
一致地测量接触角或铺展情况。
A detergent changes the system, so record concentration rather than treating all liquids as equivalent.
洗涤剂会改变体系，因此应记录浓度，而非将所有液体视为等效。
Known: zero point one zero zero kilograms of water warms by five point zero kelvin. Specific heat capacity is four thousand one hundred and eighty joules per kilogram kelvin.
已知：0.100 kg 的水升温 5.0 K；比热容为 4,180 J/(kg·K)。
Q equals m c delta T, which is zero point one zero zero times four thousand one hundred and eighty times five point zero, equaling two thousand and ninety joules.
Q = mcΔT = 0.100 × 4,180 × 5.0 = 2,090 J。
This calculation excludes energy absorbed by the container and lost to the surroundings.
此计算未包含容器吸收的能量及散失到环境中的热量。
Your turn.
轮到你了。
Use c equals 4,180 joules per kilogram per kelvin.
取比热容 c 等于每千克每开尔文 4180 焦耳。
Calculate Q for 0.200 kilograms of water heated through 5.0 kelvin.
计算 0.200 千克水升高 5.0 开尔文所需的热量 Q。
Hydrogen bonds between molecules differ from covalent bonds within a water molecule.
分子间的氢键与水分子内的共价键不同。
Polarity alone does not make every substance dissolve well in water.
仅凭极性并不能使所有物质都易溶于水。
Not true: Boiling water normally breaks its O–H covalent bonds.
不正确。 沸腾水通常不会破坏其O-H共价键。
Boiling separates molecules from each other, overcoming intermolecular hydrogen bonds.
沸腾是将分子彼此分离，克服的是分子间氢键。
The covalent O-H bonds inside each molecule survive - the steam is still H2O.
每个分子内部的O-H共价键得以保留，产生的水蒸气仍然是H2O。
Water molecules are polar.
水分子是极性的。
The same intermolecular interactions help explain surface tension and the energy needed to change temperature.
相同的分子间作用力有助于解释表面张力以及改变温度所需的能量。
Watch for hydrogen bonds between molecules differ from covalent bonds within a water molecule.
请注意：分子间的氢键与水分子内部的共价键不同。
0.100 kg of water is heated through 4.0 K.
0.100 kg的水被加热，温度升高了4.0 K。
Use 4180 J per kilogram per kelvin. Boiling separates molecules.
使用比热容为每千克每开尔文4180 J。
It does not break the covalent O-H bonds.
沸腾使分子分离，但不会破坏共价O-H键。
The result is a heat transfer of 1672 joules.
结果是传热1672焦耳。
Oxygen pulls the shared electrons, so it is slightly negative and the hydrogens slightly positive.
氧原子吸引共用电子，因此带部分负电，氢原子带部分正电。
Opposite partial charges on neighbouring molecules attract: a hydrogen bond.
相邻分子之间相反的部分电荷互相吸引，这就是氢键。
Cohesion and adhesion follow from these bonds.
内聚力和黏附力都来自这些氢键。
A fictional 0.200 kg water sample gains 4,180 J but the observed rise is 4.0 K.
一个虚构的0.200千克水样吸收了4180焦耳的热量，但实际观测到的温升为4.0开尔文。
Compare with the no-loss prediction and identify what the discrepancy can and cannot show.
将其与无热损失的理论预测值进行比较，并分析这一差异能够说明什么、不能说明什么。
The ideal rise is Q over m c, which is five point zero kelvin.
理想温升等于 Q 除以 mc，即 5.0 开尔文。
The observed rise of four point zero kelvin is lower; container heating or loss can explain it.
实际观测到的 4.0 开尔文温升较低，容器吸热或热量散失可以解释这一现象。
It does not prove which contribution dominates without measurements.
若无测量数据，则无法确定哪种因素占主导地位。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。