Question

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what was the initial temperature of the copper piece? Specific heat of copper = 0.39 J/g °C 322 °C 345 °C 356 °C 364 °C

Answer 1

Answer:

364 °C.

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 = 400.0 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 = 42.0°C - 24.0°C = 18.0°C).

∴ Q = m.c.ΔT = (400.0 g)(4.18 J/g.°C)(18.0°C) = 30096 J.

• Now, the amount of heat lost by copper (Qc) = - 30096 J.

(Qc) = m.c.ΔT,

where, Qc is the amount of heat lost by substance (Qc = - 30096 J).

m is the mass of water (m = 240.0 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 = 42.0°C - initial temperature).

∴ (- 30096 J) = (240.0 g)(0.39 J/g.°C)(42.0°C - initial temperature).

∴ (42.0°C - initial temperature) = (- 30096 J)/(240.0 g)(0.39 J/g.°C) = - 321.54 °C.

∴ initial temperature = 42.0°C + 321.54°C = 363.54 °C ≅ 364 °C.