Question

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=0.20 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). \(\mathrm{g}\) ) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C}\). The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C}\). What is the mass of the water in the container, assum- ing that all the heat lost by the copper is gained by the water?

Short answer

The mass of the water in the container can be calculated using the following equation: \(m_\text{water} = \dfrac{110 \times 0.20 \times (24.9 - 82.4)}{4.18 \times (24.9 - 22.3)}\) Upon solving, we obtain the mass of the water.

Step-by-step solution

Calculate the heat lost by the copper

To find the heat lost by the copper, use the heat transfer equation: \(Q_\text{copper} = m_\text{copper} \times c_\text{copper} \times ΔT_\text{copper}\) Given, mass of copper \(m_\text{copper} = 110 \, g\), specific heat capacity of copper \(c_\text{copper} = 0.20 \ \dfrac{J}{g^{\circ}C}\), initial temperature of copper \(T_\text{initial, copper} = 82.4^{\circ} C\), final temperature of copper and water \(T_\text{final} = 24.9^{\circ} C\). The temperature change for copper is \(ΔT_\text{copper} = T_\text{final} - T_\text{initial, copper}\) Calculate the heat lost by the copper:

Set up the equation for heat gained by the water

The equation for the heat gained by the water is: \(Q_\text{water} = m_\text{water} \times c_\text{water} \times ΔT_\text{water}\) Since all the heat lost by the copper is gained by the water, we can equate the two: \(m_\text{copper} \times c_\text{copper} \times ΔT_\text{copper} = m_\text{water} \times c_\text{water} \times ΔT_\text{water}\) We are given the initial temperature of the water \(T_\text{initial, water} = 22.3^{\circ} C\), so we can find \(ΔT_\text{water} = T_\text{final} - T_\text{initial, water}\) Also, the specific heat capacity of water is \(c_\text{water} = 4.18 \ \dfrac{J}{g^{\circ}C}\). We need to find the mass of the water, \(m_\text{water}\).

Solve for the mass of the water

Plug in the known values into the equation: \(m_\text{copper} \times c_\text{copper} \times ΔT_\text{copper} = m_\text{water} \times c_\text{water} \times ΔT_\text{water}\) \((110 \, g) \times (0.20 \, \dfrac{J}{g^{\circ}C}) \times (24.9^{\circ} C - 82.4^{\circ} C) = m_\text{water} \times (4.18 \ \dfrac{J}{g^{\circ}C}) \times (24.9^{\circ} C - 22.3^{\circ} C)\) Solve for the mass of the water, \(m_\text{water}\).

Calculate the mass of the water

Calculate the mass of the water: \(m_\text{water} = \dfrac{110 \times 0.20 \times (24.9 - 82.4)}{4.18 \times (24.9 - 22.3)}\) This will give you the mass of the water in the container.

Key concepts

Specific Heat Capacity

Understanding the specific heat capacity of a substance is key to grasping the concepts of heat transfer in chemistry. Specific heat capacity, denoted by the symbol 'c', is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin). It's essentially a measure of how much thermal energy a substance can hold.

Each material has its own unique specific heat capacity. For instance, water has a high specific heat capacity, which means it can absorb a lot of heat before its temperature increases significantly. This characteristic of water explains why it's so effective at regulating temperatures in natural and engineered thermal systems. Conversely, metals like copper have much lower specific heat capacities, which allows them to heat up and cool down more quickly.

When solving problems involving heat transfer, we must consider this property because it determines how much energy will be involved when a substance changes temperature. As seen in the textbook exercise, knowing the specific heat capacity of copper is crucial to finding out how much heat it loses when its temperature drops.

Calorimetry

Calorimetry is an experimental technique used to measure the amount of heat energy transferred in a chemical reaction, physical change, or heat capacity. This is done using an instrument called a calorimeter. The process is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another.

In the context of our exercise, calorimetry would involve measuring how much heat is lost by the copper when it is placed in water and how much heat is gained by the water. The device used for such measurements is often simple, consisting of an insulated container that minimizes heat exchange with the environment, ensuring that heat transfer occurs only between the substances within.

Calorimeters can be complex devices used in sophisticated experiments, but they all serve the same purpose: to provide an enclosed environment where the heat transfer between substances can be accurately measured, allowing for the calculation of specific heat capacities or the monitoring of temperature changes due to chemical reactions.

Thermal Equilibrium

The concept of thermal equilibrium is fundamental to heat transfer in chemistry. It refers to a state where two or more objects in contact with each other do not exchange heat, because they have reached the same temperature. At this point, there is no net flow of thermal energy between the objects.

In our textbook example, when the hot copper is placed in cooler water, heat flows from the copper to the water until both reach the same temperature. This common temperature is when thermal equilibrium is established. The principle of thermal equilibrium is what allows us to use the conservation of energy in calorimetry equations.

To elucidate, the heat lost by the copper is equal to the heat gained by the water because no heat is lost to the surroundings. This assertion is staunchly supported by the calorimetry equation that sets the heat lost by one substance equal to the heat gained by another until thermal equilibrium is reached. By mastering the concept of thermal equilibrium, students can better understand and solve problems pertaining to heat transfer.