Question

A \(150.0-\mathrm{g}\) sample of copper is heated to \(89.3^{\circ} \mathrm{C}\). The copper is then placed in \(125.0 \mathrm{~g}\) of water held in a calorimeter at \(22.5^{\circ} \mathrm{C}\). The final temperature of the mixture is \(29.0^{\circ} \mathrm{C}\). Assuming no heat is lost from the water, what is the specific heat of copper?

Short answer

The specific heat of copper is calculated using the rearranged heat formula and the known values. The exact value will be obtained after performing the calculation in Step 4.

Step-by-step solution

Calculate the heat gained by water

First, we need to find the heat gained by water using the formula \(q = mc\Delta T\). Substituting the given values, we have water's mass m as 125.0 g, the specific heat of water c as \(4.18 \, J/g^{\circ}C\), and \(\Delta T\) as the difference between the final and initial temperatures of the water, which is \(29.0^{\circ} C - 22.5^{\circ} C = 6.5^{\circ}C\). Thus, the heat gained by water is \(q = (125.0 \, g) \times (4.18 \, J/g^{\circ}C) \times (6.5^{\circ} C)\).

Calculate the heat lost by copper

Since the heat lost by the copper is equal to the heat gained by the water in this insulated system, the heat lost by copper is the same as the one calculated in Step 1. As there is no heat loss in the system, the heat content remains constant.

Calculate the specific heat of copper

Now, we have to calculate the specific heat of copper using the formula \(q = mc\Delta T\). From Step 2, we already have the value of \(q\), the mass of copper \(m\) is given as 150.0 g, and its \(\Delta T\) as \(89.3^{\circ} C - 29.0^{\circ} C = 60.3^{\circ}C\). Rearranging the formula, we get the specific heat \(c\) as \(c = \frac{q}{m\Delta T}\). Substitute the known values into this formula to find the specific heat of copper.

Substitute and calculate

Substituting the known values into the formula obtained in Step 3 will give us the specific heat of copper.

Key concepts

Calorimetry

Calorimetry is the science of measuring the amount of heat transferred to or from a substance during a chemical or physical change. It often involves the use of a calorimeter, a device designed to observe and measure this heat transfer. The key principle of calorimetry is that heat lost by one substance is equal to the heat gained by another.

• When conducting a calorimetry experiment, it's crucial to assume no heat supply or heat losses to the environment unless otherwise stated, to maintain energy conservation within the system.
• The fundamental equation used in calorimetry is: \[q = mc\Delta T\]where \(q\) is the heat in Joules, \(m\) is the mass of the substance in grams, \(c\) is the specific heat capacity in J/g°C, and \(\Delta T\) is the change in temperature in Celsius.

In this exercise, the calorimeter contains water, and we analyze the heat exchange between copper and water to determine copper's specific heat. This kind of experiment is helpful in real-life applications to identify specific material properties.

Heat Transfer

Heat transfer is the process through which thermal energy flows from one object to another, typically moving from warmer bodies to cooler ones. This principle is evident in the exercise where heated copper is placed into cooler water.

• There are three methods of heat transfer: conduction (through direct contact), convection (through fluid movement), and radiation (through electromagnetic waves).
• In this scenario, conduction is the primary method as the copper and water come directly into contact within the calorimeter, allowing heat to pass between them.

Understanding how heat transfers help in analyzing temperature changes and therefore energy exchange. The equation \(q = mc\Delta T\) reflects this movement of energy as it allows quantification of how much energy is being transferred due to a temperature difference.

Copper

Copper is a metal commonly known for its excellent electrical conductivity. It also serves as a prime example when studying thermal properties like specific heat because it's easy to measure and often used in various heat-related applications.

• Copper's specific heat might appear small compared to water (around 0.385 J/g°C), which means it warms up and cools down relatively quickly with less energy.
• In metallurgical processes and appliance design, knowing copper's specific heat helps in predicting how it will behave under thermal process conditions.

By placing heated copper in water, its specific heat can be measured using calorimetry, showing how much energy it releases as it reaches equilibrium with the water. This specific heat measurement helps in applications ranging from designing efficient heat exchangers to electrical engineering, where thermal management is essential for promoting device longevity and safety.