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A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing \(751 \mathrm{g}\) of water (density \(=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C} .\) What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{\mathrm{Cu}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\)

Short Answer

Expert verified
The final temperature of the system is approximately 6.2°C.

Step by step solution

01

Identify Known Values

We have the following information:- Mass of copper (\(m_{\text{Cu}}\)) = \(192\, \text{g}\)- Initial temperature of copper (\(T_{\text{Cu,i}}\)) = \(100.0\, ^{\circ}\text{C}\)- Specific heat capacity of copper (\(C_{\text{Cu}}\)) = \(0.385\, \text{J/g\cdot K}\)- Mass of water (\(m_{\text{w}}\)) = \(751\, \text{g}\)- Initial temperature of water (\(T_{\text{w,i}}\)) = \(4.0\, ^{\circ}\text{C}\)- Specific heat capacity of water (\(C_{\text{w}}\)) = \(4.18\, \text{J/g\cdot K}\)
02

Set Up Heat Balance Equation

At thermal equilibrium, the heat lost by copper equals the heat gained by water:\[-m_{\text{Cu}} \cdot C_{\text{Cu}} \cdot (T_{\text{f}} - T_{\text{Cu,i}}) = m_{\text{w}} \cdot C_{\text{w}} \cdot (T_{\text{f}} - T_{\text{w,i}})\]where \(T_{\text{f}}\) is the final equilibrium temperature.
03

Solve for the Final Temperature

1. Plug in the known values: \[-192 \cdot 0.385 \cdot (T_{\text{f}} - 100) = 751 \cdot 4.18 \cdot (T_{\text{f}} - 4)\]2. Simplify and solve for \(T_{\text{f}}\): - Left Side: \[-73.92(T_{\text{f}} - 100) = -73.92T_{\text{f}} + 7392\] - Right Side: \[3135.18(T_{\text{f}} - 4) = 3135.18T_{\text{f}} - 12540.72\]3. Rearrange and solve:\[-73.92T_{\text{f}} + 7392 = 3135.18T_{\text{f}} - 12540.72\]4. Combine like terms: \[7392 + 12540.72 = 3135.18T_{\text{f}} + 73.92T_{\text{f}}\]5. Solve for \(T_{\text{f}}\):\[T_{\text{f}} = \frac{19932.72}{3209.1} \approx 6.2^{\circ}\text{C}\]

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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 the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin). It is a property that tells us how much heat energy a material can store. Different substances have different specific heat capacities.
For instance, water has a high specific heat capacity of 4.18 J/g·K. This means it can absorb a lot of heat before it exhibits a noticeable change in temperature. Copper, on the other hand, has a specific heat capacity of 0.385 J/g·K, indicating it heats up and cools down quickly with smaller amounts of heat transfer.
Understanding specific heat capacity allows us to predict how different materials will react when heated or cooled, which is crucial for processes like calorimetry and understanding thermal equilibrium.
Heat Transfer
Heat transfer is the process by which thermal energy moves from a hotter entity to a cooler one. There are three primary methods of heat transfer: conduction, convection, and radiation. In our scenario involving copper and water, the focus is on conduction – the direct transfer of heat through a material.
When the hot copper is immersed in cooler water, the copper loses heat, and the water gains it until both reach thermal equilibrium. This means the heat given off by the copper equals the heat absorbed by the water, and they subsequently attain the same temperature. This transfer is governed by the law of energy conservation, which states that energy cannot be created or destroyed, only converted from one form to another. Thus, precise calculations involving heat balances are crucial to solving such problems.
Calorimetry
Calorimetry is the science of measuring the amount of heat involved in a chemical reaction or other process. The main tool for this measurement is a calorimeter, which consists of a well-insulated container. The goal is to measure changes in temperature caused by a particular reaction or physical transformation.
By employing the principles of calorimetry, we can precisely calculate the heat exchange in our copper-water situation. The formula used, relates the mass, specific heat capacity, and temperature changes of the substances involved. Here, the equation reflects heat lost by copper as equal to the heat gained by water, facilitating determination of their final shared temperature.
Through calorimetry, we effectively harness these measurements to understand and predict the thermodynamic behavior during interactions, making it a foundational tool in physical science.
  • Provides insights into energy exchange
  • Utilizes equations to relate materials' properties
  • Ensures measurement accuracy with well-constructed calorimeters

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Most popular questions from this chapter

Calculate the quantity of energy required to convert \(60.1 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) at \(0.0^{\circ} \mathrm{C}\) to \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100.0^{\circ} \mathrm{C} .\) The enthalpy of fusion of ice at \(0^{\circ} \mathrm{C}\) is \(333 \mathrm{J} / \mathrm{g}\); the enthalpy of vaporization of liquid water at \(100^{\circ} \mathrm{C}\) is \(2256 \mathrm{J} / \mathrm{g}.\)

You want to heat the air in your house with natural gas \(\left(\mathrm{CH}_{4}\right) .\) Assume your house has \(275 \mathrm{m}^{2}\) (about \(2800 \mathrm{ft}^{2}\) ) of floor area and that the ceilings are 2.50 m from the floors. The air in the house has a molar heat capacity of \(29.1 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} .\) (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{g} / \mathrm{L} .\). What mass of methane do you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C} ?\)

The specific heat capacity of copper metal is \(0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) How much energy is required to heat \(168 \mathrm{g}\) of copper from \(-12.2^{\circ} \mathrm{C}\) to \(+25.6^{\circ} \mathrm{C} ?\)

Suppose that an inch \((2.54 \mathrm{cm})\) of rain falls over a square mile of ground \(\left(2.59 \times 10^{6} \mathrm{m}^{2}\right) .\) (Density of water is \(1.0 \mathrm{g} / \mathrm{cm}^{3} .\) ) The enthalpy of vaporization of water at \(25^{\circ} \mathrm{C}\) is \(44.0 \mathrm{kJ} /\) mol. How much energy is transferred as heat to the surroundings from the condensation of water vapor in forming this quantity of liquid water? (The huge number tells you how much energy is "stored" in water vapor and why we think of storms as such great forces of energy in nature. It is interesting to compare this result with the energy given off, \(4.2 \times 10^{6} \mathrm{kJ},\) when a ton of dynamite explodes.)

Which of the following are state functions? (a) the volume of a balloon (b) the time it takes to drive from your home to your college or university (c) the temperature of the water in a coffee cup (d) the potential energy of a ball held in your hand

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