115.58 g ≅ 115.6 g.
Explanation:
Knowing that:
Heat lost by copper (Qc) = Heat gained by the water (Qw)
,
- (Qc) = (Qw).
We can calculate the amount of heat (Qw) gained by water using the relation:
Qw = m.c.ΔT,
where, Qw is the amount of heat released to water (Q = ??? J).
m is the mass of water (m = 100.0 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 24.6°C - 19.1°C = 5.5°C).
∴ Q = m.c.ΔT = (100.0 g)(4.18 J/g.°C)(5.5°C) = 2299 J.
Now, the amount of heat lost by copper (Qc) = - 2299 J.
(Qc) = m.c.ΔT,
where, Qc is the amount of heat lost by substance (Qc = - 2299 J).
m is the mass of water (m = ??? g).
c is the specific heat capacity of solution (c = 0.39 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 24.6°C - 75.6°C = - 51.0°C).
∴ (- 2299 J) = (m)(0.39 J/g.°C)(- 51.0°C).
∴ m of copper = (- 2299 J)/(0.39 J/g.°C)(- 51.0°C) = 115.58 g ≅ 115.6 g.
