17.15 解决办法的热量

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  溶液热热

  

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  How do you make solutions safely?
  ::你如何安全地解决问题?

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  When preparing dilutions of concentrated sulfuric , the directions usually call for adding the acid slowly to water with a lot of stirring. When this acid is mixed with water, a great deal of is released in the . If water were added to acid, the water would quickly heat and splatter, causing harm to the person making the solution.
  ::当准备浓缩硫的稀释时, 方向通常要求缓慢地将酸添加到水中, 并引起很多干扰。 当酸与水混合时, 大量释放在水中。 如果水添加到酸中, 水会迅速发热和溢出, 给溶液者造成伤害 。

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  Heat of Solution
  ::溶液热热

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  changes also occur when a solute undergoes the physical process of dissolving into a solvent . Hot packs and cold packs (see Figure ) use this property. Many hot packs use calcium chloride, which releases heat when it dissolves according to the equation below.
  ::当溶液进入溶剂溶解的物理过程时,也会发生变化。热袋和冷袋(见图)使用这一属性。许多热袋使用氯化钙,氯化钙根据以下方程式溶解时会释放热量。

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  $$\begin{array}{r}{\text{CaCl}}_2(s)\to {\text{Ca}}^{2+}(aq)+2{\text{Cl}}^{-}(aq)+82.8\text{ kJ}\end{array}$$

  
  ::CaCl2(s)Ca2+(aq)+2Cl-(aq)+82.8千焦耳

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  The molar heat of solution   $\begin{array}{r}(\mathrm{\Delta }{H}_{\text{soln}})\end{array}$ of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. For calcium chloride, $\begin{array}{r}\mathrm{\Delta }{H}_{\text{soln}}=-82.8\text{ kJ/mol}\end{array}$ .
  ::一种物质的摩尔溶液热(Hsoln)是当一种物质在水中溶解时吸收或释放的热。

  

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  Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.
  ::许多冷袋使用硝酸铵,硝酸铵溶解后吸收周围的热量。

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  $$\begin{array}{r}{\text{NH}}_4{\text{NO}}_3(s)+25.7\text{ kJ}\to {\text{NH}}_4^{+}(aq)+{\text{NO}}_3^{-}(aq)\end{array}$$

  
  ::NH4NO3+25.7 kJNH4+(aq)+NO3+(aq)

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  Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps to limit swelling. For ammonium nitrate, $\begin{array}{r}\mathrm{\Delta }{H}_{\text{soln}}=25.7\text{ kJ/mol}\end{array}$ .
  ::冷包通常用于治疗肌肉株状和痛关节,冷包被激活并应用于受影响地区,当硝酸铵溶解时,它吸收人体的热量,有助于限制膨胀。对于硝酸铵,Hsoln=25.7 kJ/mol。

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  Sample Problem: Heat of Solution
  ::抽样问题:解决方案的热量

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  The molar heat of solution, $\begin{array}{r}\mathrm{\Delta }{H}_{\text{soln}}\end{array}$ , of NaOH is -44.51 kJ/mol. In a certain , 50.0 g of NaOH is completely dissolved in 1.000 L of 20.0°C water in a foam cup calorimeter . Assuming no heat loss, calculate the final temperature of the water.
  ::纳奥赫的溶液的摩尔热(Hsoln)为 -44.51 kJ/mol,在一定的量的纳奥赫中,有50.0克的纳奥赫在1 000升、20.0°C的水中完全溶解,在泡沫杯卡路里计。假设没有热损失,则计算水的最后温度。

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

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  Known
  ::已知已知

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  • mass NaOH = 50.0 g
    ::质量 NaOH = 50.0克
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  • NaOH = 40.00 g/mol
    ::NaOH = 40.00克/摩尔
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  • $\begin{array}{r}\mathrm{\Delta }{H}_{\text{soln}}(\text{NaOH})=-44.51\text{ kJ/mol}\end{array}$
    ::@Hsoln(NaOH) @44.51 kJ/mol
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  • mass H ₂ O = 1.000 kg = 1000 g (assumes density = 1.00 g/mL)
    ::质量 H2O = 1.000公斤 = 1000 克(组装密度= 1.00 克/毫升)
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  • $\begin{array}{r}{T}_{\text{initial}}({\text{H}}_2\text{O})={20.0}^{\circ }\text{C}\end{array}$
    ::Tdim( H2O) = 2.0.0 C
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  • $\begin{array}{r}{c}_p({\text{H}}_2\text{O})=4.18{\text{ J/g}}^{\circ }\text{C}\end{array}$
    :H2O)=4.18 J/gC

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  Unknown
  ::未知

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  • $\begin{array}{r}{T}_{\text{final}}{\text{ of H}}_2\text{O}=?{}^{\circ }\text{C}\end{array}$
    ::H2O=决赛?

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  This is a multiple-step problem: 1) the grams NaOH is converted to moles; 2) the moles is multiplied by the molar heat of solution; 3) the joules of heat released in the dissolving process is used with the specific heat equation and the total mass of the solution to calculate the $\begin{array}{r}\mathrm{\Delta }T\end{array}$ ; 4) the  $\begin{array}{r}{T}_{\text{final}}\end{array}$ is determined from $\begin{array}{r}\mathrm{\Delta }T\end{array}$ .
  ::这是一个多步问题:1)NaOH克转换成摩尔;2)摩尔乘以溶液的摩尔热;3)溶解过程中释放的热的焦耳与特定的热方程式和计算T的溶液总质量一起使用;4)Tfinal是由T确定的。

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  Step 2: Solve .
  ::步骤2:解决。

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  $$\begin{array}{rl}& 50.0\text{ g NaOH}\times \frac{1\text{ mol NaOH}}{40.00\text{ g NaOH}}\times \frac{-44.51\text{ kJ}}{1\text{ mol NaOH}}\times \frac{1000\text{ J}}{1\text{ kJ}}=-5.56\times {10}^4\text{ J}\\ & \mathrm{\Delta }T=\frac{\mathrm{\Delta }H}{c_p\times m}=\frac{-5.56\times {10}^4\text{ J}}{4.18{\text{ J/g}}^{\circ }\text{C}\times 1050\text{ g}}=-{12.7}^{\circ }\text{C}\\ & T_{\text{final}}={20.0}^{\circ }\text{C}-(-{12.7}^{\circ }\text{C})={32.7}^{\circ }\text{C}\end{array}$$

  
  ::50.0g NaOHxx1 mol NaOH40.00 g NaOH*44.51 kJ1 mol NaOHx1000 J1 kJ*5.56×104 JTHcp×m5.56×104 J4.18 J/gC×1050 g12.7CT final=20.0C-(12.7C)=32.7C

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

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  The dissolving process releases a large amount of heat, which causes the temperature of the solution to rise. Care must be taken when preparing concentrated solutions of sodium hydroxide because of the large amounts of heat released.
  ::溶解过程释放出大量热量,导致溶液温度上升,在准备氢氧化钠的集中溶液时必须小心,因为大量释放热量。

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  Summary
  ::摘要

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  • Molar heat of solution is defined.
    ::确定溶液的摩尔热。
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  • Sample calculations using molar heat of solution are given.
    ::给出了使用摩尔热溶液的样本计算方法。

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  Review
  ::回顾

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  1. Does NaOH in solution warm or cool the water?
     ::溶液是纳奥赫热水还是冷水?
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  2. How can you tell whether a material will produce an increase or decrease in heat when dissolved?
     ::如何判断一种材料溶解后会增加或减少热量?
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  3. The sample problem was done at 20°C. Would the temperature increase be the same if the sample was run at 72°C?
     ::样本问题是在20°C时发生的。 如果样本运行在72°C时,温度升高是否相同?