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What mass of water at \(25.0^{\circ} \mathrm{C}\) must be allowed to come to thermal equilibrium with a \(1.85-\mathrm{kg}\) cube of aluminum initially at \(1.50 \times 10^{2 \circ} \mathrm{C}\) to lower the temperature of the aluminum to \(65.0^{\circ} \mathrm{C}\) ? Assume any water turned to steam subsequently recondenses.

Short Answer

Expert verified
By calculating the mass of water using the steps above, the answer is obtained. Note that the answer could differ slightly based on the precision of the quantities used in the calculations.

Step by step solution

01

Calculate Heat Lost by Aluminum

To calculate the heat lost by the aluminum, the specific heat formula may be used: \(Q = mc\Delta T\), where \(Q\) is the heat transferred, \(m\) is the mass (in this case, of the aluminum), \(c\) is the specific heat capacity (of aluminum), and \(\Delta T\) is the change in temperature. The specific heat capacity of aluminum is \(0.897 \, J/g^{\circ}C\), the mass of the aluminum cube is \(1850 \, g\), and the change in temperature is \(150^{\circ}C - 65^{\circ}C = 85^{\circ}C\). Plugging these values into the formula, we get \(Q = 1850 \times 0.897 \times 85\).
02

Equate the Heat Lost with Heat Gained by Water

The heat lost by the aluminum is gained by the water. This can be represented by the equation: \(Q_{\text{aluminum}} = Q_{\text{water}}\), where \(Q_{\text{aluminum}}\) is the heat lost by aluminum and \(Q_{\text{water}}\) is the heat gained by the water. The heat gained by the water can also be computed using the specific heat formula. However, since we are interested in the mass of the water, we rearrange the formula to calculate \(m\): \(m = Q/c\Delta T\). The specific heat capacity of water is \(4.184 \, J/g^{\circ}C\) and the change in temperature is \(65.0^{\circ}C - 25.0^{\circ}C = 40.0^{\circ}C\).
03

Calculate the Mass of Water

Substitute the calculated value of \(Q_{\text{aluminum}}\) and the known values for the specific heat and change in temperature of water into the rearranged specific heat formula to find the mass of water needed: \(m = Q_{\text{aluminum}}/4.184 \times 40\).

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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 a key concept in thermodynamics. It is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. This is a crucial property because it tells us how a material reacts to heat.
Imagine you want to heat a cup of water on a stove. The specific heat capacity of water is high, at 4.184 J/g°C. This means it takes a lot of energy to raise the temperature of water by 1°C.
In our exercise, we used the specific heat capacity to calculate the heat change in both aluminum and water.
The formula is:
  • \[ Q = mc\Delta T \]
  • Where \( Q \) is heat energy, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is temperature change.
Different materials have different specific heat capacities. For example, aluminum has a specific heat capacity of 0.897 J/g°C, much lower than water. This means aluminum will heat up and cool down more quickly than water.
Thermal Equilibrium
Thermal equilibrium is a state reached when two objects in contact no longer exchange heat. This happens when they both reach the same temperature. It's like when you leave a hot cup of coffee in a room: eventually, it will cool down to room temperature.
In this exercise, we allowed the aluminum cube and water to reach thermal equilibrium. Initially, the aluminum was hotter than the water. Because thermal equilibrium means there is no net heat flow, energy lost by the aluminum equals energy gained by the water.
Achieving thermal equilibrium is important in thermodynamics and real-world applications. For instance, engineers use it to design systems that need to maintain a specific temperature, like refrigerators or central heating systems.
Heat Transfer
Heat transfer is the movement of thermal energy from one object to another, often resulting in changes in temperature. Heat can be transferred in three main ways: conduction, convection, and radiation. However, in our exercise, we focus on conduction due to direct contact between aluminum and water.
In this situation, heat from the aluminum cube is transferred to the cooler water until both reach the same temperature.
This process is important in many scientific and engineering applications. Understanding heat transfer allows us to control and utilize energy efficiently, such as designing insulating materials or cooking food evenly.
When calculating heat transfer, we measure the amount of heat lost or gained using the specific heat formula mentioned earlier. Knowing how heat moves helps in understanding phenomena like climate change, where heat distribution in the Earth's atmosphere is a crucial factor.

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Most popular questions from this chapter

GP Into a \(0.500-\mathrm{kg}\) aluminum container at \(20.0^{\circ} \mathrm{C}\). is placed \(6.00 \mathrm{~kg}\) of ethyl alcohol at \(30.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~kg}\) ice at \(-10.0^{\circ} \mathrm{C}\). Assume the system is insulated from its environment. (a) Identify all five thermal energy uransfers that occur as the system goes to a final equilibrium temperature \(T\). Use the form "substance at \(\mathrm{X}^{\circ} \mathrm{C}\) to substance at \(Y^{\circ} \mathrm{C}_{.}^{\prime \prime}(\mathrm{b})\) Construct a table similar to the table in Example \(11.6 .\) (c) Sum all terms in the right-most column of the table and set the sum equal to zero. (d) Substitute information from the table into the equation found in part (c) and solve for the final equilibrium temperature, \(T\).

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.

18 Overall, \(80 \%\) of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation. evaporation of sweat (2 \(430 \mathrm{~kg} / \mathrm{kg})\), evaporation from the hangs \((38 \mathrm{~kJ} / \mathrm{h})\), conduction, and convection. A person working out in a gym has a metabolic rate of \(2500 \mathrm{~kJ} / \mathrm{h} .\) His body temperature is \(37^{\circ} \mathrm{C}\), and the outside temperature \(24^{\circ} \mathrm{C}\). Assume the skin has an area of \(2.0 \mathrm{~m}^{2}\) and cmissivity of \(0.97\). (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates \(0.40 \mathrm{~kg}\) of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) AL what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

A \(200-\mathrm{g}\) aluminum cup contains \(800 \mathrm{~g}\) of water in thermal equilibrium with the cup at \(80^{\circ} \mathrm{C}\). The combination of cup and water is cooled uniformly so that the temperature decreases by \(1.5^{\circ} \mathrm{C}\) per minute. At what rate is energy being removed? Express your answer in watts.

The highest recorded waterfall in the world is found at Angel Falls in Venczuela. Its longest single waterfall has a height of \(807 \mathrm{~m}\). If water at the top of the falls is at \(15.0^{\circ} \mathrm{C}\). what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising the water's temperature.

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