Chapter 9: Problem 61
We submerge a 31.1-g wafer of pure gold initially at 69.3 C into 64.2 g of water at 27.8 C in an insulated container. What is the final temperature of both substances at thermal equilibrium?
Short Answer
Expert verified
After solving the linear equation in Step 5, the final temperature \( T_f \) is the temperature at which the water and the gold reach thermal equilibrium.
Step by step solution
01
- Understanding the Concept of Thermal Equilibrium
In an insulated system, when two substances at different temperatures are put together, heat will flow from the hotter substance to the colder one until thermal equilibrium is reached. At this point, both substances will have the same final temperature. The heat gained by the colder substance is equal to the heat lost by the hotter substance.
02
- Calculating Heat Transfer
Use the formula for heat transfer, which is \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Set the heat lost by gold equal to the heat gained by water to find the equilibrium temperature.
03
- Setting Up the Equations
For gold: \( Q_{gold} = m_{gold}c_{gold}\Delta T_{gold} \). For water: \( Q_{water} = m_{water}c_{water}\Delta T_{water} \). Since \( Q_{gold} = -Q_{water} \), we have \( m_{gold}c_{gold}(T_f - T_{initial,gold}) = - m_{water}c_{water}(T_f - T_{initial,water}) \), where \( T_f \) is the final temperature.
04
- Plug in Known Values
For gold, \( m_{gold} = 31.1 g \), \( c_{gold} = 0.129 \frac{J}{g\cdot\degree C} \), \( T_{initial,gold} = 69.3\degree C \). For water, \( m_{water} = 64.2 g \), \( c_{water} = 4.186 \frac{J}{g\cdot\degree C} \), \( T_{initial,water} = 27.8\degree C \). Substitute these values into the equation from Step 3.
05
- Solve for the Final Temperature
Calculate the final temperature \( T_f \) using the equation: \( 31.1\,g \times 0.129\,\frac{J}{g\cdot\degree C} \times (T_f - 69.3\degree C) = - 64.2\,g \times 4.186\,\frac{J}{g\cdot\degree C} \times (T_f - 27.8\degree C) \). This is a linear equation in one variable, which can be solved algebraically.
06
- Checking the Results
Ensure that the calculated final temperature is between the initial temperatures of the gold and the water, as it should physically be the case. If it's not, revisit the calculations for errors.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Transfer Calculation
To understand how we calculate heat transfer, first, think of it as energy in motion between substances because of a temperature difference. The equation we use is an embodiment of the principle of conservation of energy and is given by \( Q = mc\Delta T \). Here, \(|Q| \) represents the amount of heat absorbed or released, \(|m| \) is the mass of the substance, \(|c| \) is its specific heat capacity – a measure of how much energy is needed to raise the temperature of one unit of mass of that substance by one degree – and \(|\Delta T| \) is the change in temperature.
For thermal equilibrium problems, we set the heat lost by the hot substance equal to the heat gained by the cold one because the total heat energy in an isolated system must remain constant. In our gold and water example, heat transferred to the water (gaining energy) is equal in magnitude to the heat lost by the gold (losing energy), but opposite in sign.
For thermal equilibrium problems, we set the heat lost by the hot substance equal to the heat gained by the cold one because the total heat energy in an isolated system must remain constant. In our gold and water example, heat transferred to the water (gaining energy) is equal in magnitude to the heat lost by the gold (losing energy), but opposite in sign.
Specific Heat Capacity
The specific heat capacity \( (c) \) is a key concept in thermodynamics. It represents the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). Different substances have different capacities for storing heat; for instance, water has a high specific heat capacity, which is why it's excellent for cooling systems. Gold's specific heat capacity is much lower, indicating it requires less energy to change its temperature.
In calculations, the specific heat capacity serves as a multiplier that tells us how the temperature of a substance will respond to a given amount of heat. It essentially 'translates' heat energy into temperature change. Higher specific heat capacity values mean that the substance will heat up or cool down more slowly than those with lower values, assuming the same amount of heat transfer.
In calculations, the specific heat capacity serves as a multiplier that tells us how the temperature of a substance will respond to a given amount of heat. It essentially 'translates' heat energy into temperature change. Higher specific heat capacity values mean that the substance will heat up or cool down more slowly than those with lower values, assuming the same amount of heat transfer.
Solving Thermal Equilibrium Problems
Solving thermal equilibrium problems involves finding a common final temperature when two or more substances come into thermal contact within an insulated system. The steps typically include establishing the heat transfer equation for each substance involved, ensuring the principle of conservation of energy is satisfied (heat lost equals heat gained), substituting known values, and solving the resulting equations.
Here's a practical tip: once you've calculated the final temperature, it's always a good idea to check if it logically falls between the initial temperatures of the substances involved. This common-sense check can often catch calculation mistakes. Also, pay close attention to the units used; they must be consistent across all elements of the equation for the math to work out correctly.
Here's a practical tip: once you've calculated the final temperature, it's always a good idea to check if it logically falls between the initial temperatures of the substances involved. This common-sense check can often catch calculation mistakes. Also, pay close attention to the units used; they must be consistent across all elements of the equation for the math to work out correctly.