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A \(2.50 \mathrm{~kg}\) lump of aluminum is heated to \(92.0^{\circ} \mathrm{C}\) and then dropped into \(8.00 \mathrm{~kg}\) of water at \(5.00^{\circ} \mathrm{C}\). Assuming that the lump-water system is thermally isolated, what is the system's equilibrium temperature?

Short Answer

Expert verified
The system's equilibrium temperature is approximately \(10.5^{\circ} \mathrm{C}\).

Step by step solution

01

Identify Given Data and Heat Transfer Equation

We know the specific heat capacity of aluminum, \(c_{Al} = 900 \, \mathrm{J/kg\cdot^{\circ}C}\), and for water, \(c_{water} = 4186 \, \mathrm{J/kg\cdot^{\circ}C}\). The mass of aluminum is \(m_{Al} = 2.50 \, \mathrm{kg}\) and its initial temperature is \(T_{Al} = 92.0^{\circ} \, \mathrm{C}\). The mass of water is \(m_{water} = 8.00 \, \mathrm{kg}\), and its initial temperature is \(T_{water} = 5.00^{\circ} \, \mathrm{C}\). We aim to find the equilibrium temperature \(T_{eq}\). The heat lost by aluminum should equal the heat gained by water:\[-m_{Al} \cdot c_{Al} \cdot (T_{eq} - T_{Al}) = m_{water} \cdot c_{water} \cdot (T_{eq} - T_{water})\].
02

Substitute Values into the Heat Transfer Equation

Substitute the given values into the equation:\[-2.50 \cdot 900 \cdot (T_{eq} - 92.0) = 8.00 \cdot 4186 \cdot (T_{eq} - 5.00)\].
03

Simplify and Rearrange the Equation

Expand both sides of the equation:\[-2250 (T_{eq} - 92.0) = 33488 (T_{eq} - 5.00)\].Simplify the expression:\[-2250T_{eq} + 207000 = 33488T_{eq} - 167440\].Combine like terms:\[207000 + 167440 = 33488T_{eq} + 2250T_{eq}\].
04

Solve for the Equilibrium Temperature

Combine the equilibrium temperature terms:\[374440 = 35738T_{eq}\].Solve for \(T_{eq}\) by dividing both sides by 35738:\[T_{eq} = \frac{374440}{35738} \approx 10.5^{\circ} \mathrm{C}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is an important concept when it comes to understanding heat transfer. It is defined as the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius. This value varies for different materials. For instance, aluminum has a specific heat capacity of 900 J/kg·°C, while water has a much higher specific heat capacity of 4186 J/kg·°C.

This difference in specific heat capacities means that water can absorb more heat for a given temperature change compared to aluminum. When heating or cooling methods involve multiple materials, these specific heat capacities play a crucial role in determining how the temperature of each material changes over time. The larger the specific heat capacity, the more heat the material can store, which in turn affects how the material will adjust to thermal energy changes.
Equilibrium Temperature
When two objects come into thermal contact, they eventually reach a common temperature, known as the equilibrium temperature. In a thermally isolated system, such as the lump of aluminum and water in this example, the heat loss from the hotter object (aluminum) is equal to the heat gain of the cooler object (water).

Mathematically, this is expressed as:
  • The heat lost by aluminum is given by \(-m_{Al} \cdot c_{Al} \cdot (T_{eq} - T_{Al})\).
  • The heat gained by water is given by \(m_{water} \cdot c_{water} \cdot (T_{eq} - T_{water})\).
The equilibrium temperature can be found by setting these two expressions equal to each other and solving for \(T_{eq}\).

In practices like warming, cooling, or mixing fluids and solids, calculating the equilibrium temperature helps us predict the final temperature of mixed systems, ensuring energy conservation.
Thermal Isolation
Thermal isolation is a condition whereby no heat is exchanged with the environment outside a given system. This is crucial in calculations like finding equilibrium temperature because it ensures all heat transfer occurs only between the objects in the system.

In our example, the lump-aluminum and water system is said to be thermally isolated. This means no heat is lost to the external environment and all the heat from the aluminum is transferred to the water until they reach equilibrium. Such isolation can be achieved using a well-insulated container or by conducting the experiment quickly before heat is lost.

Thermal isolation simplifies the math, as it guarantees that whatever heat one component of the system loses is gained by the other. This concept helps in applying the law of conservation of energy effectively, making it possible to accurately calculate things like equilibrium temperature in multi-material systems.

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Most popular questions from this chapter

Leidenfrost effect. \(\mathrm{A}\) water drop will last about \(1 \mathrm{~s}\) on a hot skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\). However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance \(L\) in Fig. \(18-48)\). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (a) At what rate is energy conducted from the skillet to the drop through the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

An insulated Thermos contains \(130 \mathrm{~cm}^{3}\) of hot coffee at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

70 In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloudy days, \(1.00 \times 10^{6} \mathrm{kcal}\) is needed to maintain the inside of the house at \(22.0^{\circ} \mathrm{C}\). Assuming that the water in the barrels is at \(50.0^{\circ} \mathrm{C}\) and that the water has a density of \(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\), what volume of water is required?

A \(150 \mathrm{~g}\) copper bowl contains \(220 \mathrm{~g}\) of water, both at \(20.0^{\circ} \mathrm{C}\). A very hot \(300 \mathrm{~g}\) copper cylinder is dropped into the water, causing the water to boil, with \(5.00 \mathrm{~g}\) being converted to steam. The final temperature of the system is \(100^{\circ} \mathrm{C}\). Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder?

Nonmetric version: (a) How long does a \(2.0 \times 10^{5} \mathrm{Btu} / \mathrm{h}\) water heater take to raise the temperature of 40 gal of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: (b) How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

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