95.0mLmL of H2O\rm H_2O is initially at room temperature (22.0∘C\rm{} ^{\circ}C). A chilled steel rod at 2.0∘C\rm ^{\circ}C is placed in the water. If the final temperature of the system is 21.3∘C^{\circ}C, what is the mass of the steel bar?

Question

Specific heat of water = 4.18 J/g⋅∘C\rm J/g{\cdot}^{\circ}C

Specific heat of steel = 0.452 J/g⋅∘C

Answer 1

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 mass rod

Answer 2

31.86