Question

A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what was the initial temperature of the copper piece?
Specific heat of copper = 0.39 J/g °C

A. 436 °C
B. 456 °C
C. 481 °C
D. 487 °C

Answer 1

Answer:

C. 481 °C.

Explanation:

• At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

• To find the amount of heat, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

∵ Q of copper = Q of water

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

So, the right choice is: C. 481 °C.