A piece of iron (mass = 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? The specific heat capacity of iron is 0.449 J/g·K and the specific heat capacity of water is 4.18 J/g·K.

Use the formula for heat transfer: \( q = m \cdot c \cdot \Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Set up the equation using the given values: \( m_{iron} \cdot c_{iron} \cdot (T_{final} - T_{initial, iron}) = -m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water}) \). Substitute \( m_{iron} = 25.0 \) g, \( c_{iron} = 0.449 \) J/g·K, \( T_{initial, iron} = 398 \) K, \( m_{water} = 25.0 \) g (since 25.0 mL of water is approximately 25.0 g), \( c_{water} = 4.18 \) J/g·K, and \( T_{initial, water} = 298 \) K.