Question

A 10.540 g iron bar is warmed to 250.0 c and then plunged into 5000 ml of water at 25 c. After the iron cools the final temp of the iron and water is 27.5 C. calculate the specific heat capacity and the molar heat capacity of iron.

Answer 1

Cp of iron = ?

Cp of water = 4.184 J / g oC

heat lost by iron = heat gain by water

( m Cp dT) iron = (m Cp dT) water

10.540 x Cp x (250 -27.5) = 5000 x 4.184 x (27.5 -25 )

Cp = 22.3 J / g oC

specific heat capacity = 22.3 J / g ^oC

molar heat capacity = specific heat capacity x molar mass

                                = 22.3 x 55.85 J / mole .oC

                               = 1245 J / mol. ^oC

molar heat capacity = 1245 J / mol. ^oC

note :

if anything you got wrong check your data once. because we won't get this much specific heat and molar heat for iron. if it goes wrong means the problem is only on data . based on data i solved this problem before checking all things in my solution dont try to thumbs down