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: \(77.4^{\circ} \mathrm{C}\)

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

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

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)

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.