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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 17: Temperature and HeatProblema 39
Capitolo 17, Problema 39

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

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1
Identify the specific heat capacities of the materials involved: copper, water, and iron. These values are typically found in physics textbooks or reliable online sources.
Set up the heat transfer equation based on the principle of conservation of energy, which states that the heat lost by the iron will be equal to the heat gained by the copper pot and the water. Use the formula: \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
Express the heat lost by the iron as \( Q_{iron} = m_{iron}c_{iron}(T_{initial, iron} - T_{final}) \), where \( T_{initial, iron} \) is the initial temperature of the iron block and \( T_{final} \) is the final equilibrium temperature of the system.
Express the heat gained by the copper pot and water as \( Q_{copper} = m_{copper}c_{copper}(T_{final} - T_{initial, copper}) \) and \( Q_{water} = m_{water}c_{water}(T_{final} - T_{initial, water}) \), where \( T_{initial, copper} \) and \( T_{initial, water} \) are the initial temperatures of the copper pot and water, respectively.
Set the sum of the heat gained by the copper pot and water equal to the heat lost by the iron: \( Q_{iron} = Q_{copper} + Q_{water} \). Solve this equation for \( T_{final} \) to find the final temperature of the system.

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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 crucial in this problem as it determines how much heat each material (copper, water, and iron) will absorb or release as they reach thermal equilibrium. Different materials have different specific heat capacities, affecting the final temperature calculation.
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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 this context, it implies that the total heat lost by the iron block must equal the total heat gained by the copper pot and the water, assuming no heat is lost to the surroundings. This principle is essential for setting up the equation to find the final temperature.
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Conservation Of Mechanical Energy

Thermal Equilibrium

Thermal equilibrium occurs when two or more objects in thermal contact reach the same temperature and no net heat flows between them. In this problem, the final temperature of the system is the point at which the copper pot, water, and iron block have exchanged heat and reached a uniform temperature. Understanding this concept helps in setting up the equations needed to solve for the final temperature.
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Volume Thermal Expansion
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