Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at ${25.1}^{\circ }\mathrm{C}.$ A $121.0-\mathrm{g}$ block of copper metal is heated to ${100.4}^{\circ }\mathrm{C}$ by putting it in a beaker of boiling water. The specific heat of $\mathrm{Cu}(s)$ is $0.385\mathrm{J}/\mathrm{g}-\mathrm{K}$ . The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of ${30.1}^{\circ }\mathrm{C}$ (a) Determine the amount of heat, in J, lost by the copper block. (b) Determine the amount of heat gained by the water. The specific heat of water is $4.18\mathrm{J}/\mathrm{g}-\mathrm{K}$ . (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam 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 K. Calculate the heat capacity of the calorimeter in J/K. (d)What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

Short answer

The heat lost by the copper block is 3329.2 J, and the heat gained by the water is 3135.0 J. The heat capacity of the calorimeter is 38.8 J/K. If all the heat lost by the copper block were absorbed by the water, the final temperature of the system would be 30.4°C.

Step-by-step solution

(a) Heat Lost by Copper Block

First, we need to calculate the amount of heat lost by the copper block. We can use the calorimetry equation to find this: q_Cu = m_Cu * c_Cu * ΔT_Cu Here, we have m_Cu = 121.0 g, c_Cu = 0.385 J/g-K, and ΔT_Cu = T_final - T_initial = 30.1°C - 100.4°C = -70.3°C. Plug in these values to get: q_Cu = 121.0 g * 0.385 J/g-K * (-70.3 K) = -3329.2 J The heat lost by the copper block is 3329.2 J.

(b) Heat Gained by Water

Next, we need to calculate the amount of heat gained by the water. We can use the calorimetry equation again: q_H2O = m_H2O * c_H2O * ΔT_H2O Here we have m_H2O = 150.0 g, c_H2O = 4.18 J/g-K, and ΔT_H2O = T_final - T_initial = 30.1°C - 25.1°C = 5.0°C. Plug in these values to get: q_H2O = 150.0 g * 4.18 J/g-K * 5.0 K = 3135.0 J The heat gained by the water is 3135.0 J.

(c) Heat Capacity of the Calorimeter

The difference in heat lost by the copper and heat gained by the water is due to the heat absorbed by the calorimeter. We can now calculate the heat capacity of the calorimeter: Δq = q_Cu - q_H2O = -3329.2 J - 3135.0 J = -194.2 J Since the temperature increased by 5.0 K and the heat capacity is the amount of heat necessary to raise the temperature by 1 K, the heat capacity of the calorimeter can be found by: C_cal = Δq / ΔT = -194.2 J / 5.0 K = -38.8 J/K The heat capacity of the calorimeter is 38.8 J/K.

(d) Final Temperature of the System

Finally, we need to determine the final temperature if all the heat lost by the copper were absorbed by the water. We can use the calorimetry equation one more time: q_Cu_absorbed = m_H2O * c_H2O * ΔT_final Here, q_Cu_absorbed = 3329.2 J (since all heat lost by the copper is absorbed by the water), m_H2O = 150.0 g and c_H2O = 4.18 J/g-K. So we can write the equation as: 3329.2 J = 150.0 g * 4.18 J/g-K * ΔT_final Now, we can solve for ΔT_final: ΔT_final = 3329.2 J / (150.0 g * 4.18 J/g-K) = 5.3 K Then, the final temperature would be: T_final_system = T_initial_water + ΔT_final = 25.1°C + 5.3°C = 30.4°C The final temperature of the system if all the heat lost by the copper block were absorbed by the water would be 30.4°C.

Key concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in calorimetry and thermodynamics. It refers to the amount of heat required to change the temperature of a unit mass of a material by one degree Celsius (or one Kelvin). This property is intrinsic to the material and varies from one substance to another.

In the context of the exercise, the specific heat capacity of water is significantly higher than that of copper:

• Water: 4.18 J/g-K
• Copper: 0.385 J/g-K

This means that water can absorb more heat without undergoing a large increase in temperature. For example, in our exercise, the same amount of heat causes a smaller temperature change in water than in copper. Understanding specific heat capacity helps explain why materials respond differently to heat transfer.
In formulas, specific heat capacity is represented by the letter 'c'. The heat change (q) for a substance is calculated using the formula:$$q=m\cdot c\cdot \mathrm{\Delta }T$$where:

• $m$ is the mass of the substance,
• $c$ is its specific heat capacity, and
• $\mathrm{\Delta }T$ is the change in temperature.

Specific heat capacity provides a comprehensive way to quantify how efficiently a substance can store thermal energy.

Heat Transfer

Heat transfer is the process by which thermal energy moves from a warmer object to a cooler one. In our exercise, when a hot copper block is placed in cooler water, heat flows from the copper to the water, leading to changes in their temperatures until thermal equilibrium is reached.

There are three primary modes of heat transfer:

• Conduction
• Convection
• Radiation

In a calorimeter, conduction is the main mode as heat is transferred via direct contact between the copper block and the water.

The heat lost by the copper and gained by the water can be understood using the principle of conservation of energy. In an ideal, insulated system with no heat loss, the heat gained by water should equal the heat lost by copper:$$q_{Cu}=-q_{H2O}$$This principle allows us to set up and solve equations based on measurements of temperature changes and known specific heat capacities.

Understanding heat transfer is crucial when working with calorimeters, as it aids in calculating heat changes accurately and helps identify any discrepancies caused by external heat loss or system inefficiencies.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In calorimetry exercises like the one provided, we often come across the First Law of Thermodynamics, which is pivotal in analyzing heat exchanges. The First Law states that energy cannot be created or destroyed, only transferred or converted from one form to another. Thus, the total energy within a closed system remains constant. In the context of this exercise, when copper loses heat, that energy doesn't disappear — it's transferred to the water and the calorimeter:

• The energy lost by copper is absorbed by water and the calorimeter apparatus.
• Discrepancies in calculations often point to energy absorbed by parts of the system or heat loss to the surroundings.

The calculations we perform rely on this law, ensuring that the sum of energy exchanges accounts for all temperature changes observed.

It's also essential to understand terms like "thermal equilibrium," which describes the point at which objects in contact no longer exchange heat because they are at the same temperature. While thermodynamics may seem complex, its principles are present in everyday phenomena observed in heat transfer and energy transformations.