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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 15th EditionCh 17: Temperature and HeatProblema 50
Capitolo 17, Problema 50

A laboratory technician drops a 0.08500.0850-kg sample of unknown solid material, at 100.0100.0°C, into a calorimeter. The calorimeter can, initially at 19.019.0°C, is made of 0.1500.150 kg of copper and contains 0.2000.200 kg of water. The final temperature of the calorimeter can and contents is 26.126.1°C. Compute the specific heat of the sample.

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1
Identify the principle of conservation of energy, which states that the heat lost by the sample will be equal to the heat gained by the calorimeter and its contents.
Write the equation for heat transfer: \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
Calculate the heat gained by the copper calorimeter using its specific heat capacity \( c_{\text{Cu}} = 0.385 \text{ J/g°C} \). Use the formula: \( Q_{\text{Cu}} = m_{\text{Cu}} \cdot c_{\text{Cu}} \cdot (T_{\text{final}} - T_{\text{initial, Cu}}) \).
Calculate the heat gained by the water using its specific heat capacity \( c_{\text{water}} = 4.186 \text{ J/g°C} \). Use the formula: \( Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial, water}}) \).
Set up the equation for the heat lost by the sample: \( Q_{\text{sample}} = m_{\text{sample}} \cdot c_{\text{sample}} \cdot (T_{\text{initial, sample}} - T_{\text{final}}) \). Solve for \( c_{\text{sample}} \) using the equation: \( Q_{\text{sample}} = Q_{\text{Cu}} + Q_{\text{water}} \).

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Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It is a material-specific property and is crucial in determining how a substance responds to heat energy. In this problem, it helps calculate the heat absorbed or released by the unknown material.
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Specific Heat & Temperature Changes

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In calorimetry, this principle implies that the heat lost by the hot object (the unknown sample) is equal to the heat gained by the cooler objects (the calorimeter and water).
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Conservation Of Mechanical Energy

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It involves using a calorimeter to measure the heat exchanged between substances. In this scenario, calorimetry is used to determine the specific heat of the unknown sample by analyzing the temperature changes in the calorimeter system.
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Solving Calorimetry Problems
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