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Question
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°c and water = 4.18 j/g°c.
348 k
308 k
287 k
325 k
388 k
Identify given values and system parameters
We have a hot piece of iron placed in cold water inside an insulated coffee cup calorimeter.
- Iron: \(m_{\text{Fe}} = 25.0\text{ g}\), \(T_{\text{initial, Fe}} = 398\text{ K}\), \(c_{\text{Fe}} = 0.449\text{ J/g}\cdot^{\circ}\text{C}\)
- Water: \(V_{\text{H}_2\text{O}} = 25.0\text{ mL}\) (with density \(1.00\text{ g/mL}\), \(m_{\text{H}_2\text{O}} = 25.0\text{ g}\)), \(T_{\text{initial, H}_2\text{O}} = 298\text{ K}\), \(c_{\text{H}_2\text{O}} = 4.18\text{ J/g}\cdot^{\circ}\text{C}\)
Set up the heat exchange equation
Using the Calorimeter Heat Capacity concept, we assume an ideal insulated system where the heat lost by the iron equals the heat gained by the water:
Substitute values and solve for final temperature
Using the Calorimeter Heat Capacity knowledge point
Match with the closest option
The calculated final temperature is approximately \(308\text{ K}\).
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- 348 K
- 308 K (Correct answer)
- 287 K
- 325 K
- 388 K