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 = 0.449 J $g ^ { - 1 }$ °C^-1 and water = 4.184 J $g ^ { - 1 }$ °C^-1.

Question

Quizplus · Accepted Answer

The heat lost by iron equals the heat gained by water. Using the formula $q = mc\Delta T$, set up the equation: $ (25.0 \, \text{g})(0.449 \, \text{J/g°C})(T_f - 398 \, \text{K}) = (25.0 \, \text{g})(4.184 \, \text{J/g°C})(T_f - 298 \, \text{K}) $. Solving for $T_f$ gives approximately 308 K.

A Piece of Iron (Mass = 25 g−1g ^ { - 1 }g−1

A Piece of Iron (Mass = 25 $g ^ { - 1 }$