17.12 熔化和固化热

章节大纲

• 选择节常规

  折叠 展开

  常规

  全部折叠 展开全部

  • 选择活动新闻通告

    

    新闻通告 讨论区
• 选择节聚变热和固化热

  折叠 展开

  聚变热和固化热

  

  播放段落

  What makes the ice melt?
  ::为什么冰会融化?

  播放段落

  Suppose you hold an ice cube in your hand. It feels cold because energy leaves your hand and enters the ice cube. What happens to the ice cube? It melts. However, the temperature during a phase change remains constant.  So the heat that is being lost by your hand does not raise the temperature of the ice above its temperature of 0°C. Rather, all the heat goes into the . Energy is absorbed during the process of changing ice into water. The water that is produced also remains at 0°C until all of the ice is melted.
  ::假设你手里握着一个冰块。 它感觉冷, 因为能量离开你的手, 进入冰块。 冰块会怎样? 它会融化。 但是, 阶段的温度变化仍然不变。 因此, 你的手失去的热量不会使冰的温度高于零°C的温度。 相反, 所有的热量都会进入。 能量在将冰变成水的过程中被吸收。 产生的水也保持在零°C, 直到所有的冰融化为止。

  播放段落

  Heats of Fusion and Solidification
  ::聚变热和固化热

  播放段落

  All solids absorb heat as they melt to become . The gain of heat in this endothermic process goes into changing the state rather than changing the temperature. The molar heat of fusion   $\begin{align*}(\Delta H_{\text{fus}})\end{align*}$ of a substance is the heat absorbed by one mole of that substance as it is converted from a solid to a liquid. Since the melting of any substance absorbs heat, it follows that the of any substance releases heat. The molar heat of solidification   $\begin{align*}(\Delta H_{\text{solid}})\end{align*}$ of a substance is the heat released by one mole of that substance as it is converted from a liquid to a solid. Since fusion and solidification of a given substance are the exact opposite processes, the numerical value of the molar heat of fusion is the same as the numerical value of the molar heat of solidification, but opposite in sign. In other words, $\begin{align*}\Delta H_{\text{fus}} =- \Delta H_{\text{solid}}\end{align*}$ . Figure shows all of the possible along with the direction of heat flow during each process.
  ::所有固体在熔化后吸收热量。在这一最终热过程中,热的增加会改变状态,而不是改变温度。物质聚变的摩尔热(Hfus)是该物质的摩尔热(Hfus)在从固体转化为液体时被该物质的摩尔吸收的热。由于任何物质的熔化会吸收热量,任何物质的摩尔热会释放热量。物质固化的摩尔热(Hmusy)是该物质的一个摩尔释放的热量,因为它从液体转化成固体。由于某种物质的聚变和固化是完全相反的过程,聚变聚的摩尔热的数值与固体化的摩尔热的数值相同,但符号相反。换句话说,“HfusHcusHmilly”。图显示所有可能的热量以及每个过程的热流方向。

  

  播放段落

  From left to right, heat is absorbed from the surroundings during melting, evaporation, and sublimation. Form right to left, heat is released to the surroundings during freezing, condensation, and deposition.
  ::从左到右,在熔化、蒸发和升华期间从周围吸收热量,从右到左,在冷冻、凝结和沉积期间向周围释放热量。

  播放段落

  Every substance has a unique value for its molar heat of fusion, depending on the amount of energy required to disrupt the intermolecular forces present in the solid. When 1 mol of ice at 0°C is converted to 1 mol of liquid water at 0°C, 6.01 kJ of heat are absorbed from the surroundings . When 1 mol of water at 0°C freezes to ice at 0°C, 6.01 kJ of heat are released into the surroundings.
  ::每种物质对于聚合的摩尔热具有独特的价值,这取决于破坏固体中存在的中间分子力量所需的能量量。当0°C时1兆瓦的冰在0°C时转化为1兆瓦的液态水时,从周围吸收6.01千焦耳的热量。当0°C时1兆瓦的水在0°C时冻为冰时,6.01千焦耳的热量被释放到周围。

  播放段落

  $$\begin{align*}& \text{H}_2\text{O}(s) \rightarrow \text{H}_2\text{O}(l) \qquad \Delta H_{\text{fus}}=6.01 \text{ kJ/mol} \\ & \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(s) \qquad \Delta H_{\text{solid}}=-6.01 \text{ kJ/mol}\end{align*}$$
  ::H2OH2O(l)Hfus=6.01 kJ/molH2O(l)H2O=HSUI=6.01 kJ/mol

  播放段落

  The molar heats of fusion and solidification of a given substance can be used to calculate the heat absorbed or released when various amounts are melted or frozen.
  ::某一物质的聚变和固化的摩尔热可以用来计算当各种物质被熔化或冷冻时吸收或释放的热量。

  播放段落

  Sample Problem Heat of Fusion
  ::聚融热问题

  播放段落

  Calculate the heat absorbed when 31.6 g of ice at 0°C is completely melted.
  ::计算当0°C时31.6克冰完全融化时吸收的热量。

  播放段落

  Step 1: List the known quantities and plan the problem .
  ::第1步:列出已知数量并规划问题。

  播放段落

  Known
  ::已知已知

  播放段落
  • mass = 31.6 g ice
    ::质量=31.6克冰
  播放段落
  • H ₂ O( s ) = 18.02 g/mol
    ::H2O(s) = 18.02克/摩尔
  播放段落
  • molar heat of fusion = 6.01 kJ/mol
    ::聚聚的摩尔热 = 6.01 kJ/摩尔

  播放段落

  Unknown
  ::未知

  播放段落
  • $\begin{align*}\Delta H=? \text{ J} \end{align*}$
    ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

  播放段落

  The mass of ice is first converted to moles. This is then multiplied by the conversion factor of  $\begin{align*}\left(\frac{6.01 \text{ kJ}}{1 \text{ mol}}\right)\end{align*}$ in order to find the kJ of heat absorbed.
  ::冰的质量首先转换为摩尔,然后乘以(6.01 kJ1 mol)的换算系数,以找到吸收热量的 kJ。

  播放段落

  Step 2: Solve .
  ::步骤2:解决。

  播放段落

  $$\begin{align*}31.6 \text{ g ice} \times \frac{1 \text{ mol ice}}{18.02 \text{ g ice}} \times \frac{6.01 \text{ kJ}}{1 \text{ mol ice}}=10.5 \text{ kJ}\end{align*}$$
  ::31.6克冰x1摩尔冰18.02克冰x6.01克J1摩尔冰=10.5千焦耳

  播放段落

  Step 3: Think about your result .
  ::步骤3:想想你的结果。

  播放段落

  The given quantity is a bit less than 2 moles of ice, and so just less than 12 kJ of heat is absorbed by the melting process.
  ::给定的数量略小于2摩尔的冰,因此熔化过程吸收的热量略小于12千焦耳。

   

   

  播放段落

  Summary
  ::摘要

  播放段落
  • Molar heats of fusion and solidification are defined.
    ::对聚变和固化的摩尔热作了定义。
  播放段落
  • Calculations of heat changes during fusion and solidification are described.
    ::说明在聚变和固化过程中热变化的计算。

  播放段落

  Review
  ::回顾

  播放段落
  1. In the transition from liquid to solid, is energy absorbed or released?
     ::在从液体向固体的过渡中,能源是吸收的还是释放的?
  播放段落
  2. In the transition from solid to liquid, is energy absorbed or released?
     ::在从固体到液体的过渡中,能源是吸收的还是释放的?
  播放段落
  3. How much energy is released when one mole of water at 0°C changes from liquid to solid?
     ::当0°C时一介水从液体变成固体时,能释放多少能量?