Question

A \(4-\mathrm{m} \times 5-\mathrm{m} \times 6-\mathrm{m}\) 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 \(24-\mathrm{h}\) 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 \(\mathrm{C}\)

Short answer

Answer: The minimum temperature of the water is 77.4°C.

Step-by-step solution

Calculate the total energy loss

The room loses energy at a rate of \(10,000 \mathrm{~kJ} / \mathrm{h}\) and we want to maintain the temperature for a \(24-\mathrm{h}\) period. Therefore, the total energy loss is: Total energy loss (Q) = Energy loss rate \(\times\) Time \(Q = \ 10,000 \mathrm{~kJ} / \mathrm{h} \times 24 \mathrm{~h} = 240,000 \mathrm{~kJ}\)

Calculate the energy loss in terms of water's specific heat

Now that we know the total energy loss, we can use the specific heat of water to determine the minimum temperature of the water. The specific heat of liquid water is approximately \(4.18 \mathrm{~kJ} / (\mathrm{kg} \cdot \mathrm{C})\) We can calculate the temperature difference (\(\Delta T\)) by dividing the total energy loss by the product of mass (\(m\)), specific heat (\(C_p\)) and water's mass: \(\Delta T = \frac{Q}{m \times C_p} = \frac{240,000 \mathrm{~kJ}}{1000 \mathrm{~kg} \times 4.18 \mathrm{~kJ} / (\mathrm{kg} \cdot \mathrm{C})} = 57.4^{\circ} \mathrm{C}\)

Calculate the minimum temperature of water

To determine the minimum initial temperature of the water, we will add the temperature difference to the initial temperature of the room: Minimum temperature of water = Room temperature + Temperature difference \(T_{min} = 20^{\circ} \mathrm{C} + 57.4^{\circ} \mathrm{C} = 77.4^{\circ} \mathrm{C}\) Therefore, the minimum temperature of the water when it is first brought into the room in order to maintain the room's temperature for a \(24-\mathrm{h}\) period is \(77.4^{\circ} \mathrm{C}\).