18.10
量热法
当不同温度的物体彼此接触但与其他物体隔离时，它们就会达到热平衡。 阻止热量传入或传出的容器称为热量计，使用热量计进行测量称为量热法。 一般来说，这些测量会涉及热量或比热容。 术语“量热问题” 用于解决指定对象与其周围环境热隔离的任何问题。 解决量热问题的一个重要目标是，在与周围环境隔离的物体之间进行…
量热法是在特定条件下测量物理或化学过程中释放或吸收热量的科学。
量热法的原理表明，如果两个物体与外界环境隔绝，则冷物体所吸收的热量等于热物体所释放的热量，即二者之和等于零。
若将 30 克、0 摄氏度的冰块溶解在 150 克、40 摄氏度的橙汁中，当所有冰完全融化后，液体的最终温度是多少？橙汁和冰的比热容分别为 4190 J/kg·K 和 2100 J/kg·K，冰的熔化热为 334×103 J/kg。
首先，列出已知量和未知量。
现在将量热法的原理应用于果汁和冰。
重新排列各项，以求得最终温度的表达式。
代入数值计算橙汁的最终温度，结果为 21.86 摄氏度。
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Q1: What is the basic principle of calorimetry?
Calorimetry is based on the principle of energy conservation: when two bodies at different temperatures are isolated from their surroundings, the heat gained by the colder body equals the heat lost by the hotter body. This ensures their combined heat change sums to zero. This principle applies whenever objects reach thermal equilibrium in an isolated system.
Q2: What is a calorimeter and how does it work?
A calorimeter is a container designed to prevent heat transfer between the system and surroundings. It isolates objects from external thermal influences, allowing accurate measurement of heat exchange between substances inside. The calorimeter enables precise determination of heat released or absorbed during physical or chemical processes by maintaining thermal isolation.
Q3: How do you solve a calorimetry problem involving phase changes?
To solve calorimetry problems with phase changes, identify known quantities including mass, specific heat capacity, and heat of fusion or vaporization. Apply the principle that heat lost equals heat gained. Account for energy required to change phase separately from sensible heat changes. Substitute values into the energy balance equation to find the final temperature or heat involved.
Q4: Why is thermal isolation important in calorimetry measurements?
Thermal isolation ensures that all heat exchange occurs only between the objects being studied, not with the environment. Without isolation, heat would transfer to or from surroundings, making measurements inaccurate. An insulated container or system prevents this external heat transfer, allowing the principle of heat conservation to apply correctly and yield reliable results.
Q5: What role does specific heat capacity play in calorimetry calculations?
Specific heat capacity determines how much energy is required to change an object's temperature by one degree. In calorimetry, it relates heat transferred to temperature change through the equation Q = mcΔT. Different materials have different specific heat capacities, so accounting for this property is essential for accurate calculations of heat flow and final equilibrium temperatures.
Q6: How does the orange juice and ice example demonstrate calorimetry principles?
In this example, cold ice absorbs heat from warm juice until both reach thermal equilibrium. The heat lost by juice equals the heat gained by ice, accounting for both temperature change and the energy required to melt the ice. By applying conservation of energy and using known specific heats and heat of fusion, the final equilibrium temperature can be calculated precisely.
Q7: What conditions must be met for a problem to be classified as a calorimetry problem?
A calorimetry problem requires that objects be thermally isolated from their surroundings, even if no explicit calorimeter container is mentioned. The system must prevent significant heat exchange with the environment during measurement. This isolation allows the heat gained by one object to equal the heat lost by another, enabling accurate application of energy conservation principles.