heat gained by rod + heat lost by H2O = 0
[mass rod x specific heat rod x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.
A volume of 100.mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘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)
4 answers
Thank you I ended up solving it right after I posted this. I wasn't factoring in the mas of the rod.
why are you solving for temp final??? we are looking for mass of the rod...?
18.9698