Problem 136 - General Chemistry: Principles and Modern Applications

Question

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Step-by-Step Solution

Answer

After calculations, the final temperature \( T_f \) is approximately \( 50.2^{\circ} \mathrm{C} \).

Step 1: Calculate the amount of glycerol

Firstly, determine the mass of glycerol using the volume and density. The formula is \( \text{mass} = \text{density} \times \text{volume} \). In this case, the mass of glycerol, \( m_{\text{gly}} \), is \( 1.26 \, \text{g/mL} \times 325 \, \text{mL} = 409.5 \, \text{g} \).

Step 2: Calculate the specific heat of glycerol

The specific heat of glycerol, \( C_{\text{gly}} \), is given by \( C_{\text{gly}} = C_{n} / M \), where \( C_{n} \) is the molar heat capacity and \( M \) is the molar mass of glycerol. For glycerol, \( M = 92.09 \, \text{g/mol} \). So, \( C_{\text{gly}} = 219 \, \text{JK}^{-1} \text{mol}^{-1} / 92.09 \, \text{g/mol} = 2.38 \, \text{J/g°C} \).

Step 3: Calculate the final temperature

Using the law of conservation of energy, the amount of heat gained by the glycerol is equal to the amount of heat lost by the iron. For glycerol, \( Q_{\text{gly}} = m_{\text{gly}} \cdot C_{\text{gly}} \cdot (T_f - T_{\text{gly, initial}}) \) and for iron, \( Q_{\text{Fe}} = m_{\text{Fe}} \cdot C_{\text{Fe}} \cdot (T_{\text{Fe, initial}} - T_f) \), where \( T_f \) is the final temperature. Equating the two and solving, \( T_f = (m_{\text{Fe}} \cdot C_{\text{Fe}} \cdot T_{\text{Fe, initial}} + m_{\text{gly}} \cdot C_{\text{gly}} \cdot T_{\text{gly, initial}}) / (m_{\text{Fe}} \cdot C_{\text{Fe}} + m_{\text{gly}} \cdot C_{\text{gly}}) \). Substituting values in, \( T_f = (1.22 \times 10^{3} \times 0.45 \times 99.8 + 409.5 \times 2.38 \times 26.2) / (1.22 \times 10^{3} \times 0.45 + 409.5 \times 2.38) \)

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. It can be viewed as a value that represents the 'thermal inertia' of a material; the higher the specific heat capacity, the more energy it takes to change the temperature. This is a critical concept when calculating how different materials interact thermally.

For instance, in the provided exercise, the specific heat capacity of glycerol plays a pivotal role in determining how the final temperature will be affected when it comes into contact with the hot iron piece. If glycerol had a lower specific heat capacity, a smaller amount of heat transfer would result in a larger temperature change. By understanding the specific heat capacity, we can predict and calculate how temperature changes in response to energy exchange.

Heat Transfer

Heat transfer is the process by which thermal energy moves from one body or substance to another. In many real-world applications, heat transfer is a result of a temperature difference between two objects. There are three primary modes of heat transfer: conduction, convection, and radiation.

In the context of our textbook problem, we're focusing on conduction, which is heat transfer through direct contact. When the hot iron is dropped into glycerol, heat flows from the iron to the glycerol until thermal equilibrium is reached, meaning both substances arrive at the same temperature. Calculating heat transfer allows us to use the principle of energy conservation to find out the final temperatures of the substances involved.

Law of Conservation of Energy

The law of conservation of energy states that energy cannot be created or destroyed in an isolated system; it can only be transformed from one form to another. When applying this law to thermal processes, the energy lost by the hotter object must equal the energy gained by the cooler object if no heat is lost to the surroundings.

In the exercise, the iron loses thermal energy, which is gained by the glycerol. By setting the heat lost by the iron equal to the heat gained by the glycerol, we can derive an equation that represents the conservation of energy in the system. This equation allows us to solve for the unknown final temperature that the iron and glycerol will equilibrate to.

Final Temperature Determination

Final temperature determination in a system involving two substances reaching thermal equilibrium involves using the masses, specific heat capacities, and initial temperatures of the substances. The formula that is commonly used is based on the principle that the heat lost by the hotter body is equal to the heat gained by the cooler body. This concept is illustrated in the step-by-step solution to the textbook problem.

By setting up an equation where the total heat lost by the iron equals the total heat gained by the glycerol, you solve for the final temperature, which provides a quantifiable understanding of the energy exchange between the two substances. With careful consideration of the specific heats and masses involved, the precise equilibrium temperature can be predicted, providing key insights into the thermal dynamics at play.