A piece of iron (mass = 25.00 g) at 398.0 K is placed in a styrofoam coffee cup containing 25.00 mL of water at 298.0 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 = 0.4490 J/g°C and water = 4.180 J/g°C.

When iron comes into contact with water, it gives it heat to increase its temperature. The heat that the iron yields to match the water temperature is.

Q = m . C . ΔT

Q = 25 g . 0,4990 J/g°C (298K - 398K)

Q = 25 g . 0,4990 J/g°C - 100°C = -1247,5 J

Kelvin or Celsius is the same, because ultimately the formula only takes the result of a subtraction

This is the heat (-1247,5 J) that water wins to increase its temperature. As in iron it is a detached heat, the result was negative, however, for water it is a gained heat, that's why it is positive.

1247,5 J = 25g . 4,180 J/g°C (Tfinal - 25°C)

For units, it's more confortable to work in Celsius.

1247,5 J / (25g . 4,180 J/g°C) = Tfinal - 25°C

11,93°C = Tfinal - 25°C

11,93°C + 25°C = Tfinal = 36,93 °C

It is reasonable that the water has increased its temperature because the iron temperature was 100 degrees higher, before both substances came into contact