A volume of 125mL of H 2 O is initially at room temperature (22.0 ∘ C ). A chilled steel rod at 2.00 ∘ C is placed in the water. If the final temperature of the system is 21.4 ∘ C , what is the mass of the steel bar?
Use the following values:

specific heat of water = 4.18 J/(g⋅ ∘ C)

specific heat of steel = 0.452 J/(g⋅ ∘ C)

Question

A volume of 125mL  of H 2 O  is initially at room temperature (22.0  ∘ C ). A chilled steel rod at 2.00  ∘ C  is placed in the water. If the final temperature of the system is 21.4  ∘ C  , what is the mass of the steel bar? 
Use the following values:

specific heat of water = 4.18 J/(g⋅ ∘ C) 

specific heat of steel = 0.452 J/(g⋅ ∘ C)

anonymous · Answer

The energy lost by the water = the energy gained by the steel rod, so:

-q = q

-mc(Tf-Ti) = mc(Tf-Ti)

Just plug the values in and solve for mass of steel bar.

anonymous · Answer

Remember to change mL -> grams.  Easy conversion.