Show the calculation of the final temperature for a 20.8 gram piece of iron heated to 100oc which has been added to a 55.3 gram sample of water at 25.3oc in a coffee cup calorimeter.
c (water) = 4.184 j/g oc; c (fe) = 0.449 j/g oc

Final Temperature = 28.2 oC

Explanation:

Information given;

Mass of Iron = 20.8g

Initial Temperature of Iron = 100C

Mass of water = 55.3g

Initial temperature of water = 25.3 C

The presence of a coffee cup calorimeter hints that there is no heat loss to the surrounding and that the iron and water are at thermal equilibrium.

Thermal equilibrium means that there is no heat transfer going on between the bodies, which simply means that the bodies are at the same temperature.

Hence, both bodies would the same final temperature (T2)

H = M * C * ΔT (For iron)

H = 20.8 * 0.449 * ( 100 - T2)

H = 9.3392 ( 100 - T2)

H = 933.92 - 9.3392T2

H = M * C * ΔT (For water)

H = 55.3 * 4.184 * (T2 - 25.3)

H = 231.3752 (T2 - 25.3)

H = 231.3752T2 - 5853.79

Since they are in thermal equilibrium it means H (Iron) = H (water).

This leads to;

933.92 - 9.3392T2 = 231.3752T2 - 5853.79

231.3752T2 +  9.3392T2  = 5853.79 + 933.92

240.7144 T2 = 6787.71

T2 = 28.2 oC