# Problem: 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.

###### FREE Expert Solution

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:

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

q = heat, J

+qabsorbs heat
–qloses heat

m = mass (g)
c = specific heat capacity = J/(g·°C)
ΔT = Tf – Ti = (°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
Tf aluminum = ?
ccopper = 0.88 J/g°C

mass water = 180 g
Ti water = 85
°C
Tf water = ?
cwater = 4.184 J/(g
°C)

80% (479 ratings) ###### Problem Details

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.