In this problem, we explore the principles of calorimetry by examining the cooling of a hot iron chunk through the application of water. The scenario involves a 1.6 kg piece of iron initially at 600 °C, which is cooled down to 130 °C by dripping water at 20 °C over it. The water absorbs heat from the iron, causing it to boil away completely.

To analyze the heat transfer, we set up a calorimetry equation where the heat gained by the water equals the negative heat lost by the iron. The water undergoes two distinct processes: first, it heats up to its boiling point, and then it undergoes a phase change from liquid to vapor. The heat absorbed by the water can be expressed as:

Q_w = Q_w1 + Q_w2 = m_wc_wΔT_w + m_wL_v

Here, m_w is the mass of the water, c_w is the specific heat capacity of water (4186 J/kg·°C), ΔT_w is the change in temperature of the water, and L_v is the latent heat of vaporization (2.256 × 10⁶ J/kg).

On the other side of the equation, the heat lost by the iron is given by:

Q_iron = -m_ic_iΔT_i

Where m_i is the mass of the iron (1.6 kg), c_i is the specific heat capacity of iron (470 J/kg·°C), and ΔT_i is the change in temperature of the iron (from 600 °C to 130 °C).

By substituting the known values into the calorimetry equation, we can solve for the mass of the water:

m_w(c_wΔT_w + L_v) = -m_ic_iΔT_i

Calculating the temperature changes, we find:

ΔT_w = 100 °C - 20 °C = 80 °C

ΔT_i = 130 °C - 600 °C = -470 °C

Substituting these values into the equation allows us to isolate m_w and ultimately calculate its value. The result shows that approximately 0.136 kg of water is required to cool the iron effectively. This illustrates the high specific heat capacity of water, which enables it to absorb significant amounts of heat with relatively small mass changes.

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