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)

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.

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...?

To find the mass of the steel bar, we can apply the principle of energy conservation. The energy lost by the steel rod must be gained by the water. This can be determined using the formula:

Q = m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Where:
Q is the energy transferred from the steel rod to the water (in Joules)
m1 is the mass of the steel rod (in grams)
c1 is the specific heat of steel (in J/(g⋅°C))
ΔT1 is the change in temperature of the steel rod (final temperature - initial temperature) (in °C)
m2 is the mass of the water (in grams)
c2 is the specific heat of water (in J/(g⋅°C))
ΔT2 is the change in temperature of the water (final temperature - initial temperature) (in °C)

From the given information, we have:
c1 = 0.452 J/(g⋅°C)
c2 = 4.18 J/(g⋅°C)
ΔT1 = 21.50 °C - 2.00 °C = 19.50 °C
ΔT2 = 21.50 °C - 22.00 °C = -0.50 °C

Substituting these values into the equation, we have:
Q = m1 * 0.452 * 19.50 = m2 * 4.18 * (-0.50)

Simplifying the equation, we get:
m1 = (m2 * 4.18 * -0.50) / (0.452 * 19.50)

Now, we can solve for m1, the mass of the steel bar.

18.9698