SPECIFIC HEAT AND HEAT OF FUSION

HEAT EXCHANGE: When two or more objects at different temperatures are brought together in an isolated environment, they eventually reach the same temperature by the process of heat exchange. That is, warmer materials transfer heat to colder materials until their temperatures are the same. The energy which is being transferred is referred to as internal energy, energy associated with random molecular motion on the microscopic scale. This energy can be divided into kinetic energy and the potential energy arising from the intermolecular attractive forces in the material. If we have two isolated substances and assume no loss to the environment, then the conservation of energy principle implies that the energy lost by one substance must be gained by the other. Letting Q stand for the quantity of heat, this idea may be expressed as

热交换：当两个或多个温度不同的物体被放在一个孤立的环境中时，它们最终会通过热交换的过程达到相同的温度。也就是说，较热的物质会将热量传递给较冷的物质，直到它们的温度相同。正在传递的这种能量被称为内能，这是与微观尺度上分子随机运动相关的能量。这种能量可以分为动能和由于材料中分子间吸引力产生的势能。如果我们有两件孤立的物质，并假设没有能量损失到环境中，那么能量守恒原理表明，一个物质失去的能量必须由另一个物质获得。令Q表示热量的量，这个想法可以表示为

Temperature is a measure of the average kinetic energy of the random molecular motion. As a body gains or loses kinetic energy its temperature will increase or decrease. The temperature change is symbolized by DT where

温度是随机分子运动平均动能的度量。当一个物体获得或失去动能时，其温度将升高或降低。温度的变化用ΔT表示，其中

Although the internal energy possessed by an object is directly proportional to its mass, it does not follow that two objects of the same mass and temperature have the same amount of internal energy. Temperature reflects only the kinetic energy portion of the internal energy, so a substance with a greater fraction of its internal energy in the form of potential energy will have a greater internal energy at a given temperature. Thus a gram of water at 50°C will have a much larger internal energy than a gram of copper at 50°C and it will take a much larger energy input to heat the water to 60°C than to produce the same temperature change in copper. This property is reflected in the quantity called the specific heat (S). The specific heat is defined as the quantity of heat required to raise the temperature of 1 gram of a substance 1°C.

尽管物体所具有的内能与其质量成正比，但这并不意味着质量相同、温度相同的两个物体具有相同的内能。温度只反映内能中动能的部分，因此，一种物质如果其内能中更大比例以势能形式存在，则在给定温度下其内能会更大。因此，50°C的1克水的内能将远大于50°C的1克铜的内能，并且将需要更大的能量输入来将水加热到60°C，而不是产生铜的相同温度变化。这一性质体现在称为比热容（S）的量上。比热容定义为将1克物质温度升高1°C所需热量的量。

Taking all of the above into consideration, the quantity of heat lost or gained by a body can be calculated by the relation

综合上述内容，物体吸收或放出的热量可通过以下关系式计算

其中M是物质的质量。

* Internal energy can be measured in calories. The dietary Calorie is a kilocalorie or 1000 calories.

内能可以以卡路里为单位进行测量。饮食中的卡路里是一个大卡，即1000卡路里。

LABORATORY EXERCISE: The specific heat of a metal sample will be determined. A summary of the experiment follows, and the apparatus video accessible from the right sidebar will show you the pieces of equipment involved.

实验室练习：测定金属样品的比热容。实验总结如下，右侧侧边栏中的装置视频将展示所用的仪器设备。

Heated metal is poured into an insulated container of cool water. Since care will be taken to prevent heat from entering or escaping the container, it is assumed that the amount of energy lost by the hot metal is gained by the water.

将加热的金属倒入一个隔热的冷水容器中。由于会采取措施防止热量进入或逸出容器，因此假设热金属失去的热量会由水获得。

From relationships (2) and (3) the amount of heat gained by the water can be computed. Since this must be the amount of heat lost by the metal, relationship (3) can be applied to the metal sample. When substitution of the known quantities is done, the remaining unknown quantity is the specific heat of the metal, which is then calculated from the equation.

根据关系式（2）和（3），可以计算出水吸收的热量。由于这必须是金属失去的热量，关系式（3）可以应用于金属样品。当代入已知量后，剩下的未知量是金属的比热容，然后可以通过方程计算得出。

将锅炉装满约2/3的水，立即开始加热。

将已知质量的金属放入锅炉杯中，如下所示：

确定空锅炉杯的质量

b. Fill the boiler cup about 2/3 full of the metal material provided and redetermine the mass.

b. 将锅炉杯装满约2/3的金属材料，并重新确定其质量。

3. Carefully place the boiler cup into the boiler so the metal can begin heating.

4. Using the graduated cylinder, pour 100 ml or a convenient volume of cool water which will fill the calorimeter cup about 2/3 full. Since water has a density of 1 gram/cm³ = 1 gram/ml, the number of grams equals the number of ml of water.

使用量筒，倒入约100毫升或方便的体积的冷水，使热量杯约装满2/3。由于水的密度为1克/立方厘米=1克/毫升，因此水的毫升数等于克数。

5. Place one thermometer in the boiler cup and carefully work it down into the metal particles.

将一个温度计放入锅炉杯中，小心地将其插入金属颗粒中。

6. When the metal is nearly finished heating, place another thermometer into the calorimeter and record the initial temperature of the water.

当金属几乎加热完毕时，将另一个温度计放入量热器中，并记录水的初始温度。

7. When the metal reaches about 95°C (which is to be the initial temperature of the metal), quickly remove the boiler cup from the boiler and pour the hot metal into the calorimeter. Record the initial temperature of the metal.

当金属温度达到约95°C（这将是金属的初始温度）时，迅速将锅炉杯从锅炉中取出，将热金属倒入量热计中。记录金属的初始温度。

8. Quickly put the lid on the calorimeter. Place the thermometer through the hole in the top to the lid and stir the metal and water GENTLY.

8. 快速将盖子盖在量热计上。将温度计通过顶部的孔插入盖子中，并轻轻搅拌金属和水。

9. After you have inserted the thermometer and stirred the mixture, record the temperature. This is the final temperature of the metal and the water. Record the temperature. The temperature reading should drift down slowly from this value as the mixture cools toward room temperature, so it is important to make this measurement promptly.

9. 在插入温度计并搅拌混合物后，记录温度。这是金属和水的最终温度。记录温度。温度读数应从这个值缓慢下降，随着混合物冷却到室温，因此必须尽快进行此测量。

10. Calculate the specific heat of the metal from the data you have collected. Make a tentative identification the the metal by referring to the table to specific heats provided.

根据你收集的数据计算金属的比热容。通过参考提供的比热容表，尝试确定该金属的种类。

水的熔化热第二部分

CHANGE OF PHASE-SOLID TO LIQUID: An increase in internal energy is required to convert a solid to a liquid. Conversely, a reduction in internal energy can cause liquids to freeze or solidify. These solid-liquid phase changes occur without a change in temperature, i.e., no change in average kinetic energy occurs. The heat of fusion of a substance is the heat exchange required to melt one gram of the substance (calories/gm).

相变—固态到液态：要将固态转化为液态，需要增加内能。相反，减少内能可以使液体冻结或凝固。这些固-液相变过程中温度不发生变化，即平均动能没有变化。物质的熔化热是指将1克物质熔化所需的热量（卡/克）。

In this part of the experiment, the heat of fusion of water will be determined. Warm water will be used to melt ice, and the change in temperature of the water in the calorimeter will be used to compute the amount of energy extracted from the water to melt the ice. The ice must absorb heat in order to melt. The heat absorbed can be expressed as

在本实验的这一部分，将测定水的熔化热。用温水来融化冰，水温在量热计中的变化将用来计算从水转移到融化冰的热量。冰必须吸收热量才能融化。吸收的热量可以表示为

where Lf is the symbol for the heat of fusion in calories/gram. The necessary heat will be transferred from the warm water to the ice.

其中，Lf 表示以卡/克为单位的熔化热。所需的热量将从温水传递到冰中。

1. Warm some water to about 15°C above room temperature. (Water from the hot water tap may be hot enough.)

1. 将水加热至约室温以上15°C。（从热水龙头取的水可能足够热。）

2. Mass the empty calorimeter with the styrofoam cup and record its mass. Remass it when it is approximately half full of the warm water and record the new mass. This must be done carefully, as it cannot be repeated later. Calculate the mass of the warm water.

给空 calorimeter 加上 Styrofoam 杯并记录其质量。当它大约装满一半的温水时，再次称量并记录新的质量。这必须谨慎进行，因为之后无法重复。计算温水的质量。

3. Gently stir the water and record its temperature using the 50°C thermometer.

轻轻搅拌水，并使用50°C温度计记录其温度。

4. Dry pieces of the ice provided with a piece of paper towel. Slip them into the water, stirring gently and being careful not to splash water out. Continue to stir the water while adding ice to it one piece at a time. When the temperature of the water is approximately 15°C below room temperature, stop adding ice. When the last bit of ice has melted, record the lowest temperature that the water reaches. This temperature is the final temperature of the warm water and of the ice water.

4. 将冰块与纸巾一起放入水中，轻轻搅拌并注意不要溅出水。在加入冰块时继续搅拌水。当水温大约比室温低15°C时，停止加冰。当最后一块冰融化后，记录水达到的最低温度。这个温度是温水和冰水的最终温度。

5. Mass the calorimeter and its contents and determine the mass of ice that was added.

称量量热计及其内容物，并确定已加入的冰的质量。

6. Calculate the heat of fusion and find the percent deviation from the accepted value 79.72 cal/gm.

6. 计算熔化热，并找出与公认值79.72 cal/g的百分比偏差

要使10克水升温20°C，需要多少卡路里？

2. What temperature change would the above amount of energy cause in 10 grams of lead?

上述能量会使10克铅的温度变化多少？

3. How would the results for the heat of fusion of water have been affected if the ice were not dried before being placed in the calorimeter? Would this have made the measured heat of fusion larger or smaller?

如果在将冰放入量热计之前没有将其干燥，那么水的熔化热结果会如何变化？这会使测得的熔化热更大还是更小？