Question

A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.

Short answer

Answer: The minimum amount of water needed to cool the iron to 23.0°C is approximately 12.71 kg.

Step-by-step solution

Identifying the Given Information

We are given the following information: - Mass of iron (\(m_1\)) = 0.38 kg - Initial temperature of iron (\(T_{i1}\)) = \(498^{\circ} \mathrm{C}\) - Final temperature of iron (\(T_{f1}\)) = \(23^{\circ} \mathrm{C}\) - Initial temperature of water (\(T_{i2}\)) = \(20^{\circ} \mathrm{C}\) - Final temperature of water (\(T_{f2}\)) = \(23^{\circ} \mathrm{C}\) We need to find the minimum amount of water (mass of water, \(m_2\)). The specific heat capacities of iron and water are: - Specific heat capacity of iron (\(c_1\))= 450 J/kg\(\cdot\)K - Specific heat capacity of water (\(c_2\))= 4190 J/kg\(\cdot\)K

Writing the Heat Transfer Equation

Since heat lost by iron is equal to heat gained by water, we can write the equation: \(m_1 \cdot c_1 \cdot (T_{f1} - T_{i1}) = m_2 \cdot c_2 \cdot (T_{f2} - T_{i2})\)

Calculating the Change in Temperature

First, let's find the change in temperature for both iron and water: - Change in temperature of iron (\(\Delta T_1\)) = \(T_{f1} - T_{i1}\) = \(23^{\circ} \mathrm{C} - 498^{\circ} \mathrm{C} = -475^{\circ} \mathrm{C}\) - Change in temperature of water (\(\Delta T_2\)) = \(T_{f2} - T_{i2}\) = \(23^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 3^{\circ} \mathrm{C}\)

Plugging Values into the Heat Transfer Equation

Now we can plug all the values into the heat transfer equation: \((0.38\,\text{kg}) \cdot (450\,\text{J/kg}\cdot\text{K}) \cdot(-475\,\text{K}) = m_2 \cdot (4190\,\text{J/kg}\cdot\text{K}) \cdot (3\,\text{K})\)

Solving for the Minimum Mass of Water (\(m_2\))

Solve the equation to find the minimum mass of water: \(m_2 = \dfrac{(0.38\,\text{kg}) \cdot (450\,\text{J/kg}\cdot\text{K}) \cdot(-475\,\text{K})}{(4190\,\text{J/kg}\cdot\text{K}) \cdot (3\,\text{K})}\) \(m_2 \approx 12.71\,\text{kg}\) Therefore, the minimum amount of water needed to cool the iron to \(23.0^{\circ} \mathrm{C}\) is approximately 12.71 kg.