Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22°C. If the heat released during condensation is used only to warm the iron block, what is the final temperature (in °C) of the iron block? (Assume a constant enthalpy of vaporization for water of 2260 J/g and specific heat capacity of iron as 0.449 J/g°C.)

• A. 24.3°C
• B. 25.0°C
• C. 23.5°C
• D. 22.8°C

Answer: (A)

Verified step by step guidance

1. Calculate the heat released by the condensation of water using the formula: \( q = m \times \Delta H_{vap} \), where \( m \) is the mass of the water (0.95 g) and \( \Delta H_{vap} \) is the enthalpy of vaporization (2260 J/g).
2. Determine the amount of heat absorbed by the iron block using the formula: \( q = m \times c \times \Delta T \), where \( m \) is the mass of the iron (75.0 g), \( c \) is the specific heat capacity of iron (0.449 J/g°C), and \( \Delta T \) is the change in temperature.
3. Set the heat released by the water equal to the heat absorbed by the iron: \( m_{water} \times \Delta H_{vap} = m_{iron} \times c_{iron} \times \Delta T \).
4. Solve for \( \Delta T \) (the change in temperature of the iron block) by rearranging the equation: \( \Delta T = \frac{m_{water} \times \Delta H_{vap}}{m_{iron} \times c_{iron}} \).
5. Add the initial temperature of the iron block (22°C) to \( \Delta T \) to find the final temperature of the iron block.