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, assuming that all the heat lost by the copper is gained by the water?

Short answer

The mass of the water in the container is \(72.5\ \mathrm{g}\).

Step-by-step solution

Write down the known variables

From the exercise, we know the following: - The mass of copper (m_copper) = 110 g - The specific heat capacity of copper (c_copper) = 0.20 J/g°C - The initial temperature of copper (T_initial_copper) = 82.4°C - The initial temperature of water (T_initial_water) = 22.3°C - The final temperature of both copper and water (T_final_both) = 24.9°C

Write the equation for heat transfer

We know that when we put the hot copper inside the container, the copper will lose heat and the water will gain heat until both reach the same final temperature. So, we can use the following equation for heat transfer when all the heat lost by the copper is gained by the water: \(Q_{lost\_copper} = Q_{gained\_water}\)

Find the heat lost by copper

We can find the heat lost by the copper using the equation: \(Q_{lost\_copper} = m_{copper} \times c_{copper} \times \Delta T_{copper}\) Where: - m_copper = mass of copper - c_copper = specific heat capacity of copper - ∆T_copper = change in temperature of copper = |T_final_both - T_initial_copper| Now, substituting the values, we get: \(Q_{lost\_copper} = (110) \times (0.20) \times (|24.9 - 82.4|)\) Calculating the heat lost by copper: \(Q_{lost\_copper} = 110 \times 0.20 \times 57.5 = 1265\ \mathrm{J}\)

Find the mass of water using heat gained by water

Now, we will find the mass of water using the heat gained by water: \(Q_{gained\_water} = m_{water} \times c_{water} \times \Delta T_{water}\) Where: - m_water = mass of water - c_water = specific heat capacity of water = 4.18 J/g°C (given) - ∆T_water = change in temperature of water = |T_final_both - T_initial_water| Since we know that \(Q_{lost\_copper} = Q_{gained\_water}\), we can rewrite the equation as: \(m_{water} = \frac{Q_{gained\_water}}{c_{water} \times \Delta T_{water}}\) Substituting the values, we get: \(m_{water} = \frac{1265}{(4.18) \times (|24.9 - 22.3|)}\) Calculating the mass of water: \(m_{water} = \frac{1265}{4.18 \times 2.6} = 72.5\ \mathrm{g}\) So, the mass of the water in the container is 72.5 g.

Key concepts

Specific Heat Capacity

Understanding the concept of specific heat capacity is crucial in solving problems involving heat transfer. Specific heat capacity is a measure of the amount of thermal energy required to change the temperature of a unit mass of a substance by 1 degree Celsius. It helps us understand how different materials react to heat.

For example, copper has a relatively low specific heat capacity of 0.20 J/g°C. This means it doesn't require much energy to increase its temperature. On the other hand, water has a higher specific heat capacity of 4.18 J/g°C, indicating that it can absorb more heat without a significant change in temperature. This difference explains why water is effective in moderating temperature changes in various applications.

When applying this concept to exercises, remember:

• Low specific heat capacity means quicker temperature changes with heat addition or removal.
• High specific heat capacity means slower temperature changes with heat addition or removal.
• It's essential to know the specific heat capacity values to calculate heat transfer effectively.

Temperature Change

Temperature change is at the core of analyzing heat transfer between substances. It describes the difference between the initial and final temperatures, denoted as \(\Delta T\). In the given problem, we see temperature changes in both copper and water.

The copper starts at 82.4°C and cools down to 24.9°C, resulting in a temperature change of \(|24.9 - 82.4| = 57.5°C\). For water, the change is from 22.3°C to 24.9°C, equating to \(|24.9 - 22.3| = 2.6°C\). Understanding how to calculate these changes is crucial for solving heat transfer problems, as they dictate the amount of heat gained or lost by each substance.

When calculating temperature changes, remember:

• Always subtract the final temperature from the initial temperature to get the magnitude of change.
• Use absolute values as temperature changes cannot be negative in this context.
• Temperature change provides a direct input into calculating the energy transferred during heating/cooling processes.

Thermal Equilibrium

Thermal equilibrium is a state when two or more linked systems reach a common temperature and no net energy transfer occurs between them. This concept is essential in understanding how heat flows from hotter objects to cooler ones until equilibrium is achieved.

In the exercise, copper and water eventually reach a final temperature of 24.9°C. They achieve thermal equilibrium as the copper loses heat and the water gains it until both are at the same temperature. This equilibrium state is vital for solving such problems because it implies that the heat lost by one component is equal to the heat gained by the other.

Important points about thermal equilibrium include:

• The principle that energy flows from hot to cold until both objects are equally warm.
• The idea that net energy change (total heat transfer) is zero at equilibrium.
• Ensures conservation of energy during heat exchange, allowing accurate calculation of unknown values, like mass, in heat transfer problems.