Question

Suppose \(0.0100 \mathrm{~kg}\) of steam (at \(100.00^{\circ} \mathrm{C}\) ) is added to \(0.100 \mathrm{~kg}\) of water (initially at \(19.0^{\circ} \mathrm{C}\) ). The water is inside an aluminum cup of mass \(35.0 \mathrm{~g} .\) The cup is inside a perfectly insulated calorimetry container that prevents heat exchange with the outside environment. Find the final temperature of the water after equilibrium is reached.

Short answer

Answer: The final temperature of the water after equilibrium is reached is approximately \(22.3\,^{\circ}\mathrm{C}\).

Step-by-step solution

List the known values

We know: - Mass of steam (\(m_s\)) = \(0.0100\,\mathrm{kg}\) - Initial temperature of steam (\(T_s\)) = \(100.00\,^{\circ}\mathrm{C}\) - Mass of water (\(m_w\)) = \(0.100\,\mathrm{kg}\) - Initial temperature of water (\(T_w\)) = \(19.0\,^{\circ}\mathrm{C}\) - Mass of aluminum cup (\(m_c\)) = \(35.0\,\mathrm{g}\) = \(0.035\,\mathrm{kg}\)

Convert initial temperatures to Kelvin

To avoid negative temperatures and simplify calculations, convert the given initial temperatures to Kelvin: \(T_s = 100.00\,^{\circ}\mathrm{C} + 273.15 = 373.15\,\mathrm{K}\) \(T_w = 19.0\,^{\circ}\mathrm{C} + 273.15 = 292.15\,\mathrm{K}\)

Write down specific heat capacities

Write down the specific heat capacities of steam, water, and aluminum: - Specific heat capacity of steam (\(c_s\)) = \(2.01 \times 10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}\) - Specific heat capacity of water (\(c_w\)) = \(4.18 \times 10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}\) - Specific heat capacity of aluminum (\(c_c\)) = \(0.9 \times 10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}\)

Set up conservation of energy equation

Heat gained by water and aluminum cup is equal to heat lost by steam: \(Q_w + Q_c = Q_s\) Where: - \(Q_w = m_w \times c_w \times (T_f - T_w)\) - \(Q_c = m_c \times c_c \times (T_f - T_w)\) - \(Q_s = m_s \times c_s \times (T_s - T_f)\) - \(T_f\) is the final temperature of the system in Kelvin So: \(m_w \times c_w \times (T_f - T_w) + m_c \times c_c \times (T_f - T_w) = m_s \times c_s \times (T_s - T_f)\)

Solve for final temperature \(T_f\)

Rearrange and solve the equation for the final temperature \(T_f\): \((m_w \times c_w + m_c \times c_c) \times T_f - m_w \times c_w \times T_w - m_c \times c_c \times T_w = m_s \times c_s \times T_s - m_s \times c_s \times T_f\) \((m_w \times c_w + m_c \times c_c + m_s \times c_s) \times T_f = m_w \times c_w \times T_w + m_c \times c_c \times T_w + m_s \times c_s \times T_s\) \(T_f = \frac{m_w \times c_w \times T_w + m_c \times c_c \times T_w + m_s \times c_s \times T_s}{m_w \times c_w + m_c \times c_c + m_s \times c_s}\) Plug in the known values and solve for \(T_f\): \(T_f = \frac{0.100 \times 4.18 \times 10^3 \times 292.15 + 0.035 \times 0.9 \times 10^3 \times 292.15 + 0.0100 \times 2.01 \times 10^3 \times 373.15}{0.100 \times 4.18 \times 10^3 + 0.035 \times 0.9 \times 10^3 + 0.0100 \times 2.01 \times 10^3} = 295.45\,\mathrm{K}\)

Convert final temperature to Celsius

Convert the final temperature from Kelvin to Celsius: \(T_f = 295.45\,\mathrm{K} - 273.15 = 22.3\,^{\circ}\mathrm{C}\) The final temperature of the water after equilibrium is reached is approximately \(22.3\,^{\circ}\mathrm{C}\).

Key concepts

Specific Heat Capacity

Understanding specific heat capacity is crucial for solving problems involving temperature changes in different substances. It is defined as the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). This property is unique to each material, which means that different substances absorb or release different amounts of heat for the same temperature change.

For example, the specific heat capacity of water is relatively high, which means water can absorb a lot of heat before its temperature rises significantly. This property is exploited in exercises like our textbook problem, where the steam and water reach thermal equilibrium. We determine the heat exchange by using the formula
\( Q = m \times c \times \text{Δ}T \),where
\( m \) is the mass of the substance,
\( c \) is the specific heat capacity, and
\( \text{Δ}T \) is the change in temperature.

Thermal Equilibrium

When two or more objects at different temperatures come into contact within an isolated system, they will exchange heat energy until they reach the same temperature. This state is known as thermal equilibrium. At this point, there is no net heat flow between the objects, and their temperatures remain constant unless an external influence is added.

In the context of our exercise, steam is mixed with water in an insulated container. Here, the steam, water, and aluminum cup will eventually settle at a common temperature. Achieving thermal equilibrium is central to calorimetry exercises, where you often need to find the final uniform temperature of the system.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In the realm of heat transfer, this principle indicates that the total heat lost by hot objects will equal the total heat gained by cold objects within an isolated system.

In heat transfer calculations, we apply this principle to determine unknown variables, such as the final temperature in a system. The exercise provided is an application of the conservation of energy, as the heat released by the steam cooling (and potentially condensing) will be equal to the heat absorbed by the water and the aluminum cup until thermal equilibrium is reached.

By setting up an equation that equates the heat lost to heat gained, we can solve for the final temperature of the system, ensuring that the total energy before and after the heat exchange remains constant.