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 21.30 ∘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.mass of the steel =

Answer 1

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g