Question

A volume of 125 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 21.00  ∘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)

Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

To determine the mass of steel bar

first expect that there is no trade of warmth with the environment.

at that point ,

heat taken by the steel bar (Qs) = heat lost by water (Qw) however negative sign

Qs= Qw ---------->(1)

Consider the expression regarding the relation between the heat, temperature& mass

Q = C x (T-T0) x m substitute this in condition (1)

Cs x (T-Ts) x ms = Cw x (T-Tw) x mw

mass of water = volume x thickness (1 g/ml )

= 125 x 1 g/ml =125 g

discover mass of steel pole

ms = Cw x (T-Tw) x mw/Cs x (T-Ts)

= - 4.18 x (21-22) x 125/0.452 x( 21-2)

= - 4.18 x (- 1) x 125/0.452 x 19

= 522.5/8.588

Mass of steel bar= 60.84 g