Physics 2201 Fall 2015 Homework IV Part 2 Show All Work

Physics 2201 Fall 2015 Homework IV Part 2, Show all work and/or explain all answers

Analyze the thermal interactions within various calorimetric systems, including mixtures of water, metals, ice, and steam, considering energy conservation, heat transfer, phase changes, and equilibrium states. The problems involve calculating temperature changes, energy flow, melting of ice, and the equilibrium temperature in closed systems involving phase transitions and heat exchanges.

Paper For Above instruction

The given problems involve complex thermodynamic processes within calorimetric systems, requiring application of principles such as conservation of energy, heat transfer, latent heat during phase changes, and thermal equilibrium. They challenge understanding of how energy flows between different components, how phase changes absorb or release heat, and how to model these interactions mathematically to find equilibrium states.

Thermal Equilibrium in Metal-Water Systems

The first problem involves a well-insulated calorimeter containing water and an aluminum cup, into which a copper cylinder of known mass and initial temperature is introduced. The key to solving this problem is recognizing that, in a closed, insulated system, the total energy remains constant. The initial temperature of the water and aluminum are known, and the problem asks what the final temperature read by the thermometer will be after the copper reaches thermal equilibrium with the water and the cup.

To find this, we consider the heat transfer from the copper to the water and the aluminum cup, assuming no energy loss to the surroundings. The specific heats of copper, aluminum, and water are well-documented. The heat lost by the copper cylinder as it warms up (or cools down) equals the heat gained by the water and the aluminum cup as they reach a common final temperature. Since the system is insulated, the sum of these heat exchanges equals zero:

m_copper  c_copper  (T_initial_copper - T_final) + m_water  c_water  (T_initial_water - T_final) + m_aluminum  c_aluminum  (T_initial_aluminum - T_final) = 0

By plugging in the actual masses, specific heats, and initial temperatures, one can solve for T_final, the temperature of the system once equilibrium is established. Given the initial conditions, the arithmetic will yield a final temperature slightly above 22°C, as the cold copper cools to that temperature, or alternatively, slightly below if the copper is initially colder than the water — but in this case, since the freezer temperature is below 0°C, the copper initially being at -8°C will absorb heat from the water and aluminum until equilibrium.

Similarly, for the second part, if the entire cup, water, and copper system is removed and warmed to room temperature (22°C), the energy flow involved is the heat required to raise the system’s temperature from 0°C (or the current state) to 22°C. This energy can be calculated by summing the heat capacity of each component multiplied by the temperature change.

Melting of Ice and Heat Exchange with Water

Another major part of the problem involves a quantity of ice at 0°C and how much energy is needed to melt it completely. The latent heat of fusion for ice is a well-known constant (approximately 334 J/g). The total energy required to melt a specific mass of ice at 0°C is simply:

Q_melt = m_ice * L_fusion

Where m_ice is the mass of ice, and L_fusion is the latent heat of fusion per gram. With this, the problem extends to whether the water initially at 30°C has enough energy to supply this melting process. The heat that can be transferred from the water as it cools down from 30°C to a lower temperature (or 0°C if it cools enough) is given by:

Q_available = m_water  c_water  (T_initial_water - T_final_water)

By comparing Q_available to Q_melt, it is possible to determine if all the ice melts. If Q_available ≥ Q_melt, all ice can melt; otherwise, only part of the ice does. The resultant temperature of the system will adjust accordingly, reaching an equilibrium where no more heat transfer occurs. In cases with smaller amounts of water, such as 100 g, similar calculations determine if the ice can melt entirely, often resulting in an equilibrium temperature at or near 0°C where the system stabilizes.

To model these energy exchanges precisely, we set up an energy balance equation of the form:

ΔEsys = Q_in - Q_out = 0

with explicit terms accounting for heat gained or lost by each component, including phase changes and temperature-dependent heat transfer. The unknown equilibrium temperature is solved by substituting known quantities, latent heats, specific heats, and initial conditions, resulting in algebraic solutions for the temperature at which the net energy change is zero.

Steam Injection into the Ice-Water Mixture

The scenarios involving steam at 100°C injected into a sealed container containing ice and water involve the transfer of latent heat from condensing steam. When 100 g of steam is injected, and it condenses, it releases heat that can melt some ice or raise the temperature of the mixture. The key is calculating the amount of heat released by the condensation of steam:

Q_condensation = m_steam * L_vaporization

where L_vaporization (roughly 2260 J/g for water) is the latent heat of vaporization. If this energy exceeds the energy required to melt all the ice, then all the ice can melt, and the system will reach a new equilibrium temperature. The amount of ice melted is then determined by balancing the heat released against the heat required to melt the ice and raise the temperature of water and melted ice to the equilibrium temperature.

In the case of 5 g of steam, less energy is released (Q = 5 g * 2260 J/g ≈ 11,300 J), which may not be sufficient to melt all ice or raise the temperature significantly, depending on initial conditions. Calculations can determine the exact extent of melting and final temperature by setting up energy balance equations similar to earlier, considering the specific heats of water and ice, and the latent heat involved.

Conclusion

These problems demonstrate the importance of heat transfer principles and energy conservation in thermodynamics. By carefully accounting for phase change enthalpies, specific heats, and initial conditions, one can predict temperature outcomes and phase states in complex, insulated, or sealed systems. Setting up and solving the appropriate energy equations provides a powerful tool for analyzing thermal interactions in physical systems.

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