Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains \(150.0 \mathrm{~g}\) of water at \(25.2^{\circ} \mathrm{C}\). A \(200-\mathrm{g}\) block of silver metal is heated to \(100.5^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Ag}(s)\) is \(0.233 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})\). The \(\mathrm{Ag}\) is added to the calorimeter, and after some time the contents of the cup reach a constant temperature of \(30.2^{\circ} \mathrm{C} .(\mathbf{a})\) Determine the amount of heat, in J, lost by the silver block. (b) Determine the amount of heat gained by the water. The specific heat of water is \(4.184 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}) .(\mathbf{c})\) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam \(^{\circ}\) cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by \(1 \mathrm{~K} .\) Calculate the heat capacity of the calorimeter in \(\mathrm{J} / \mathrm{K}\). (d) What would be the final temperature of the system if all the heat lost by the silver block were absorbed by the water in the calorimeter?

Short answer

The heat lost by the silver block is \(3261.18~J\), and the heat gained by the water is \(3138~J\). The heat capacity of the calorimeter is \(24.636~J/K\). If all the heat lost by the silver block were absorbed by the water in the calorimeter, the final temperature of the system would be \(30.38^{\circ}C\).

Step-by-step solution

Find the heat lost by the silver block

To find the heat lost by the silver block, we'll use the equation \(q = mcΔT\), where \(m = 200~g\), \(c = 0.233~J/(g\cdot K)\), and \(ΔT = T_{initial} - T_{final} = 100.5^{\circ}C - 30.2^{\circ}C\). \(q_{silver} = (200)(0.233)(100.5-30.2) = 200 \times 0.233 \times 70.3~J\) Calculating the value, we get: \(q_{silver} = 3261.18~J\)

Calculate the heat gained by the water

Now, we will calculate the heat gained by water using the same equation \(q = mcΔT\), where \(m = 150~g\), \(c = 4.184~J/(g\cdot K)\), and \(ΔT = T_{final} - T_{initial} = 30.2^{\circ}C - 25.2^{\circ}C\). \(q_{water} = (150)(4.184)(30.2 - 25.2) = 150 \times 4.184 \times 5~J\) Calculating the value, we get: \(q_{water} = 3138~J\)

Calculate the heat capacity of the calorimeter

To calculate the heat capacity of the calorimeter, we will use the difference between the heat lost by the silver block and the heat gained by the water. We will also divide this difference by the total temperature change that occurred during the experiment (\(ΔT\)). \(Heat \, capacity = \frac{q_{silver} - q_{water}}{ΔT} = \frac{3261.18 - 3138}{5}~J/K\) Calculating the value, we get: \(Heat \, capacity = 24.636~J/K\)

Calculate the final temperature if all the heat lost by silver were absorbed by water

Finally, we will find the final temperature if all the heat lost by the silver block were absorbed by the water in the calorimeter. Since the heat lost by the silver block would be equal to the heat gained by the water, we can write: \(q_{silver} = mcΔT_{final}\) Solving for the final temperature change (\(ΔT_{final}\)), we get: \(ΔT_{final} = \frac{q_{silver}}{mc_{water}} = \frac{3261.18~J}{(150~g)(4.184~J/(g\cdot K))}\) Calculating the value, we get: \(ΔT_{final} = 5.18~K\) Now, we will add this temperature change to the initial temperature of water to get the final temperature of the system: \(T_{final} = T_{initial} + ΔT_{final} = 25.2^{\circ}C + 5.18~K\) Calculating the value, we get: \(T_{final} = 30.38^{\circ}C\)

Key concepts

Specific Heat

Specific heat is a property of a substance that tells you how much heat is needed to change the temperature of a certain amount of the substance. It's defined as the amount of heat per unit mass required to raise the temperature by one degree Celsius (or Kelvin). This concept is essential in calorimetry, which measures the heat of chemical reactions or physical changes.
When calculating specific heat, we use the formula: \( q = mcΔT \) where: - \( q \) is the heat absorbed or released, measured in joules (J) - \( m \) represents mass, typically in grams (g) - \( c \) is the specific heat capacity, expressed in \(J/(g\cdot K) \) - \( ΔT \) is the temperature change, \( T_{final} - T_{initial} \)
By understanding specific heat, you can determine how different materials respond to heat input or output, making it fundamental for studying thermal dynamics and energy transfer.

Heat Capacity

Heat capacity is a measure of the amount of heat required to change a system's temperature by one degree. Unlike specific heat, which is evaluated per unit mass, heat capacity considers the whole system. It's represented in units of \( J/K \).
For example, a calorimeter's heat capacity tells us how much heat the entire system (including the calorimeter itself) needs for a temperature change. In the given exercise, we calculated the calorimeter's heat capacity by determining the difference in heat lost by the silver block and gained by the water. Then, divide this by the total temperature change: \[ Heat \, capacity = \frac{q_{silver} - q_{water}}{ΔT} \]
A higher heat capacity implies that more energy is needed to change the temperature, which is an essential consideration in experiments where precision and efficiency in thermal energy transfer are critical.

Temperature Change

Temperature change, a key variable in calorimetry, refers to the difference between the final and initial temperatures of a substance experiencing heat flow. In the formula \( q = mcΔT \), \( ΔT \), the temperature change, is crucial for calculating heat transfer.
It's expressed as: \( ΔT = T_{final} - T_{initial} \)
In practice, a positive \( ΔT \) indicates that heat is gained or absorbed by the system, raising the temperature. A negative \( ΔT \) shows heat loss. For instance, in our exercise, the silver block lost heat, causing a temperature drop, while the water gained heat, resulting in a temperature rise. Understanding temperature change is vital because it allows you to quantify the energic impact on a given material or system, providing insight into its thermal equilibrium and energy requirements.

Thermal Equilibrium

Thermal equilibrium occurs when two interacting systems reach the same temperature and, consequently, no net heat flow exists between them. This concept is important in calorimetry, where the goal is often to measure how heat flows from a warmer to a cooler substance until equilibrium is achieved.
In a calorimetry experiment, thermal equilibrium allows us to observe how heat energy is distributed. For instance, in the described exercise, when the heated silver block and water in the calorimeter reached the same temperature at 30.2°C, thermal equilibrium was achieved. For calculations, it’s assumed all the heat lost by silver is equal to the heat gained by water, except for losses to the calorimeter system. Reaching thermal equilibrium means you've balanced the energy exchange, allowing precise calculations of specific heat or heat capacity. It helps in understanding the broader implications of energy conservation in thermally conductive systems.