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/gÂ°c and water = 4.18 j/gÂ°c. 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/gÂ°c and water = 4.18 j/gÂ°c. 308 k 348 k 325 k 388 k 287 k

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Chemistry

Villarreal

28 October, 07:48

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/gÂ°c and water = 4.18 j/gÂ°c. 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/gÂ°c and water = 4.18 j/gÂ°c. 308 k 348 k 325 k 388 k 287 k

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Alma Campos · Answer 1

The final temperature of the water is the equilibrium temperature, or the also the final temperature of the iron after a long period of time. Applying the conservation of energy:

m, iron*C, iron*ΔT = - m, water*C, water*ΔT

The density of water is 1000 g/mL.

(25 g) (0.449 J/g·°C) (T - 398 K) = - (25 mL) (1000 g/mL) (4.18 J/g·°C) (T - 298)

Solving for T,

T = 298.01 K