Question

The heat of hydration of dough, which is \(15 \mathrm{kJ} / \mathrm{kg}\) will raise its temperature to undesirable levels unless some cooling mechanism is utilized. A practical way of absorbing the heat of hydration is to use refrigerated water when kneading the dough. If a recipe calls for mixing \(2 \mathrm{kg}\) of flour with \(1 \mathrm{kg}\) of water, and the temperature of the city water is \(15^{\circ} \mathrm{C}\), determine the temperature to which the city water must be cooled before mixing in order for the water to absorb the entire heat of hydration when the water temperature rises to \(15^{\circ} \mathrm{C}\). Take the specific heats of the flour and the water to be 1.76 and \(4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) respectively.

Short answer

Answer: The initial temperature of the city water should be approximately 7.82°C.

Step-by-step solution

Identify the given information

First, we should list down the given information from the problem statement: - Heat of hydration of dough: \(H = 15\, \mathrm{kJ/kg}\) - Mass of flour: \(m_f = 2\, \mathrm{kg}\) - Mass of water: \(m_w = 1\, \mathrm{kg}\) - Specific heat capacity of flour: \(c_{f} = 1.76\, \mathrm{kJ/(kg\cdot^{\circ}C)}\) - Specific heat capacity of water: \(c_{w} = 4.18\, \mathrm{kJ/(kg\cdot^{\circ}C)}\) - Final temperature of water: \(T_{f,w} = 15\, ^{\circ}\mathrm{C}\)

Determine the heat absorbed by the flour

First, we will determine the heat absorbed by the flour when mixed with the water. The heat transfer can be calculated using the equation: $$Q = m\cdot c \cdot \Delta T$$ In this case, the heat absorbed by the flour (\(Q_f\)) is: $$Q_f = m_f \cdot c_f \cdot (T_{f,w} - T_i)$$ Here, \(T_i\) is the initial temperature of the flour, which is equal to the city water temperature (\(15\, ^{\circ}\mathrm{C}\)), as they are mixed together in the end.

Determine the heat absorbed by the water

Similarly, the heat absorbed by the water (\(Q_w\)) is: $$Q_w = m_w \cdot c_w \cdot (T_{f,w} - T_{i,w})$$ Here, \(T_{i,w}\) is the initial temperature of the water that we need to find.

Calculate the total heat absorption required

The water needs to absorb the entire heat of hydration, so the total heat required, \(Q_t\), can be written as: $$Q_t = H \cdot m_f + Q_f$$ By substituting the equation for \(Q_f\) from step 2, we get: $$Q_t = H \cdot m_f + m_f \cdot c_f \cdot (T_{f,w} - T_i)$$

Use conservation of energy to find the initial temperature of water

Now, we will use the conservation of energy principle, that states the total heat absorbed by the flour and the water should be equal. Therefore, we have: $$Q_t = Q_w$$ By substituting the equations for \(Q_t\) from step 4 and \(Q_w\) from step 3, we get: $$H \cdot m_f + m_f \cdot c_f \cdot (T_{f,w} - T_i) = m_w \cdot c_w \cdot (T_{f,w} - T_{i,w})$$

Solve for the initial temperature of water

Now, we will solve the equation obtained in step 5 for \(T_{i,w}\). First, substitute the given values and isolate \(T_{i,w}\): $$(15 \mathrm{kJ/kg}) \cdot (2 \mathrm{kg}) + (2 \mathrm{kg}) \cdot [1.76 \mathrm{kJ/(kg \cdot^{\circ}C)}]\cdot [(15^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C})] = (1 \mathrm{kg}) \cdot [4.18\mathrm{kJ/(kg\cdot^{\circ}C)}] \cdot [(15^{\circ}\mathrm{C} - T_{i,w})]$$ Simplify the above equation: $$30 \,\mathrm{kJ} = 4.18 \, \mathrm{kJ}/^{\circ}\mathrm{C} \cdot (15^{\circ}\mathrm{C} - T_{i,w})$$ Divide both sides by \(4.18\,\mathrm{kJ}/^{\circ}\mathrm{C}\): $$7.18\,^{\circ}\mathrm{C} = 15^{\circ}\mathrm{C} - T_{i,w}$$ Finally, solve for \(T_{i,w}\): $$T_{i,w} = 15^{\circ}\mathrm{C} - 7.18\,^{\circ}\mathrm{C}$$ $$T_{i,w} = 7.82\,^{\circ}\mathrm{C}$$ Therefore, the city water must be cooled to approximately \(7.82^{\circ}\mathrm{C}\) before mixing, in order for the water to absorb the entire heat of hydration when it rises to \(15^{\circ}\mathrm{C}\).