Question

A 4-m \(\times 5-\mathrm{m} \times 6-\mathrm{m}\) room is to be heated by one ton ( \(1000 \mathrm{~kg}\) ) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of \(10,000 \mathrm{~kJ} / \mathrm{h}\). The room is initially at \(20^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), and is maintained at an average temperature of \(20^{\circ} \mathrm{C}\) at all times. If the hot water is to meet the heating requirements of this room for a 24-h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. Answer: \(77.4^{\circ} \mathrm{C}\)

Short answer

Answer: Approximately 77.4°C.

Step-by-step solution

Calculate the heat loss for a 24-hour period

Given the average rate of heat loss as 10000 kJ/h, we need to find the total amount of heat loss in 24 hours. We can do so by multiplying the given rate by the number of hours. Heat loss = rate × hours Heat loss = 10000 kJ/h * 24 h

Calculate the energy required to maintain the room's temperature

Now that we have the heat loss, we can calculate the energy required to keep the room's temperature constant by using the heat loss value. Energy_required = Heat loss Energy_required = 10000 kJ/h * 24 h

Calculate the energy required to raise the water temperature

Next, we need to find out how much energy is required to raise the temperature of 1000 kg of water to the desired temperature. We will use the following equation: Energy_required = m * c * ΔT, where m = mass of water (1000 kg) c = specific heat capacity of water (4.18 kJ/kg°C) ΔT = Temperature difference (T_final - T_initial)

Determine the minimum initial temperature of the water

Now we will equate the energy required to heat the room (from step 2) with the energy required to raise the temperature of the water (from step 3) and solve for the final temperature of the water. Energy_required (Step 2) = Energy_required (Step 3) 10000 kJ/h * 24 h = 1000 kg * 4.18 kJ/kg°C * (T_final - T_initial) Solve the equation for T_final: T_final = (10000 kJ/h * 24 h) / (1000 kg * 4.18 kJ/kg°C) + T_initial Since the initial temperature of the room is 20°C, we can plug that value into our equation to find the final temperature of the water. T_final = (10000 kJ/h * 24 h) / (1000 kg * 4.18 kJ/kg°C) + 20°C T_final~77.4°C Thus, the minimum temperature of the water when it is first brought into the room is approximately 77.4°C.

Key concepts

Heat Transfer

Heat transfer is all about the movement of thermal energy from one area or material to another. This concept plays a crucial role in everyday phenomena, such as heating a room or cooking food. The transfer happens because of a temperature difference, meaning energy moves from a warmer object to a cooler one. There are three main modes through which heat is transferred: conduction, convection, and radiation.

• Conduction: Occurs in solids where heat moves through direct contact.
• Convection: Happens in fluids, like air and water, where heat flows through a fluid's bulk movement.
• Radiation: Involves energy transfer through electromagnetic waves and doesn't need a medium.

In the room heating exercise, the heat transfer ensures that the energy from the hot water moves into the room air, compensating for the heat lost to the outside. The goal is to maintain a stable room temperature despite the external cooling effect.

Specific Heat Capacity

Specific heat capacity is a property that signifies how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a measure of a material's ability to store and retain heat. The equation used for this purpose is:\[ q = m \times c \times \Delta T \]where:

• \(q\): Heat energy (in joules or kilojoules)
• \(m\): Mass of the substance (in kilograms)
• \(c\): Specific heat capacity (in kJ/kg°C)
• \(\Delta T\): Temperature change (in °C)

Water’s specific heat capacity is 4.18 kJ/kg°C, which is relatively high compared to many other substances. This means water can absorb and store a lot of heat without large changes in temperature. In the exercise, we use this property of water to calculate how much it can heat the room by knowing how much its temperature will change.

Temperature Difference

A crucial part of thermodynamics, temperature difference, or \(\Delta T\), is the driving force behind heat transfer. It represents the difference between the final and initial temperatures in a process and determines how energy flows from one object to another. In practice, when you have a higher temperature difference, the rate of heat transfer will be faster. That's why extremely hot water is more effective at heating a room compared to lukewarm water. In our case, we looked at maintaining the room at a constant temperature. We do this by ensuring the hot water brought into the room is at an adequate initial temperature. We calculated the temperature difference using the initial temperature of the room and the final water temperature, ensuring it compensates for the heat lost over 24 hours. This difference is what guarantees the room remains cozy even when it’s chilly outside.