Question

One ton \((1000 \mathrm{kg})\) of liquid water at \(50^{\circ} \mathrm{C}\) is brought into a well-insulated and well-sealed \(4-\mathrm{m} \times 5-\mathrm{m} \times 6-\mathrm{m}\) room initially at \(15^{\circ} \mathrm{C}\) and 95 kPa. Assuming constant specific heats for both air and water at room temperature, determine the final equilibrium temperature in the room.

Short answer

Answer: The final equilibrium temperature in the room is approximately 48.8°C.

Step-by-step solution

Calculate the volume and mass of air in the room

First, we need to find the volume of the room, using the formula: \(V_\text{room} = l × w × h\) where l, w, and h are the length, width, and height of the room. \(V_\text{room} = 4\text{m} × 5\text{m} × 6\text{m} = 120\,\text{m}^3\) Now we will calculate the mass of air in the room using the formula for an ideal gas: \(PV = m_\text{air} × R × T\) Rearranging for \(m_\text{air}\), we get: \(m_\text{air} = \frac{PV}{RT}\) where P is the pressure (95 kPa), V is the volume of the room (120 m³), R is the specific gas constant for air (287 J/kg·K), and T is the initial temperature of the room (288 K, converted from 15°C). \(m_\text{air} = \frac{95\,000\,\text{Pa} × 120\,\text{m}^3}{287\,\text{J/kg}\cdot\text{K} × 288\,\text{K}} \approx 139.9\,\text{kg}\)

Perform an energy balance by equating energy changes

We will now perform an energy balance by equating the energy taken from the water to the energy absorbed by the air. \(Q_\text{water} = m_\text{water} × c_\text{water} × (T_\text{water, initial} - T_\text{final})\) \(Q_\text{air} = m_\text{air} × c_\text{air} × (T_\text{final} - T_\text{air, initial})\) Since the room is well-insulated and well-sealed, there is no heat transfer between the room and its surroundings, and thus the energy taken from the water is equal to the energy absorbed by the air. \(Q_\text{water} = Q_\text{air}\) We can use the specific heat capacity of water (\(c_\text{water}\)) = 4,184 J/kg·K and the specific heat capacity of air (\(c_\text{air}\)) = 1,005 J/kg·K.

Solve for final temperature

Equating the energy changes and solving for the final temperature, we get: \(m_\text{water} × c_\text{water} × (T_\text{water, initial} - T_\text{final}) = m_\text{air} × c_\text{air} × (T_\text{final} - T_\text{air, initial})\) Replacing the given values and the calculated masses, we get: \(1000\,\text{kg} × 4,\!184\,\text{J/kg}\cdot\text{K} × (323\,\text{K} - T_\text{final}) = 139.9\,\text{kg} × 1,\!005\,\text{J/kg}\cdot\text{K} × (T_\text{final} - 288\,\text{K})\) Now, solve for \(T_\text{final}\): \(T_\text{final} \approx 321.8\,\text{K}\) Converting the final temperature back to Celsius, we get: \(T_\text{final} \approx 48.8^{\circ}\,\text{C}\) Thus, the final equilibrium temperature in the room is approximately 48.8°C.

Key concepts

Energy Balance Equation

In any thermodynamic process where the system is isolated from its surroundings, the energy within the system must be conserved. An energy balance equation is a mathematical expression representing this conservation of energy. In the context of our exercise, the energy transferred from the warm water must equal the energy absorbed by the cooler air in the room, as there is no energy exchange with the outside environment due to the room being well-insulated and sealed.

An energy balance can be written as:
\[ Q_\text{total} = \sum Q_\text{in} - \sum Q_\text{out} = 0 \]
Where \(Q_\text{total}\) represents the net heat transfer, \(Q_\text{in}\) is the heat entering the system, and \(Q_\text{out}\) is the heat leaving the system. When solving this kind of problem, setting up the energy balance equation is crucial to find the final equilibrium temperature.

Specific Heat Capacity

Specific heat capacity is a measure of how much energy is required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). It is an intrinsic property of the material and is often denoted by the symbol \(c\), with units of \(\text{J/kg}\cdot\text{K}\).

In our problem, we use the specific heat capacity of water and air to calculate how much energy the water loses as it cools down, and how much energy the air gains as it heats up. The formulas involved are:
\[ Q_\text{water} = m_\text{water} \times c_\text{water} \times (T_\text{water, initial} - T_\text{final}) \]
\[ Q_\text{air} = m_\text{air} \times c_\text{air} \times (T_\text{final} - T_\text{air, initial}) \]
Understanding specific heat capacity is essential because it allows us to calculate the energy involved in heating or cooling a substance without needing to measure the energy directly. In thermodynamics, this is a vital step towards understanding energy transfers in physical systems.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas with the amount of gas (in moles or mass). For our problem, it's especially helpful because it allows us to calculate the mass of the air in the room, given the pressure, temperature, and volume of the room.

The ideal gas law is often stated as:
\[ PV = nRT \]
Here, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature. When the mass of the gas is used instead of the number of moles, the equation becomes:
\[ PV = m_\text{gas} \times R_\text{specific} \times T \]
\(R_\text{specific}\) is the specific gas constant, which differs for each gas, and \(m_\text{gas}\) is the mass of the gas. By measuring the room's initial conditions and applying the ideal gas law, we can find the mass of air in the room, which is a critical step in calculating the final equilibrium temperature.