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.

• A. 340 K
• B. 320 K
• C. 310 K
• D. 330 K

Answer: (B)

Verified step by step guidance

1. Identify the principle of conservation of energy, which states that the heat lost by the iron will be equal to the heat gained by the water, assuming no heat is lost to the surroundings.
2. Write the equation for heat transfer: \( q_{iron} = -q_{water} \). This means the heat lost by iron is equal to the negative of the heat gained by water.
3. 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.
4. 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.
5. Solve the equation for \( T_{final} \), which is the final temperature of both the iron and the water. This involves algebraic manipulation to isolate \( T_{final} \) on one side of the equation.