We’re being asked to determine the temperature at which the copper piece was heated *initially*.

We will use the heat released by the copper piece to calculate its initial temperature. Recall that heat can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{\u2206}}{\mathbf{T}}}$

q = heat, J

• **+q** → **absorbs **heat

• **–q** → **l****oses **heat

m = mass (g)

c = specific heat capacity = J/(g·°C)

ΔT = T_{f} – T_{i} = (°C)

A 45 g aluminum spoon (specific heat 0.88 J/g°C) at 24°C is placed in 180 mL (180 g) of coffee at 85°C and the temperature of the two become equal.

**Given:**

mass aluminum = 45 g

Ti_{ aluminum} = 24°C

T_{f aluminum} = ?

c_{copper} = 0.88 J/g°C* *

mass water = 180 g

T_{i water} = 85°C

T_{f water} = ?

c_{water} = 4.184 J/(g∙°C)

A 45 g aluminum spoon (specific heat 0.88 J/g°C) at 24°C is placed in 180 mL (180 g) of coffee at 85°C and the temperature of the two become equal.

What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.

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