Question

Consider two Styrofoam coffee cups. If cup #1 contains 80.0 mL of water at 96.0 ^oC and cup #2 contains 60.0 mL of water at 11.0 ^oC, what is the final temperature, in ^oC, when the contents of cup #1 are poured into cup #2? Ignore the heat capacity of the cup. Assume that the density of water is 1.00 g/mL.

Answer 1

Cup #1

mass of water = volume * density

mass of water = (80.0 mL) * (1.00 g/mL)

mass of water = 80.0 g

Initial temperature = 96.0 ^oC

Cup #2

mass of water = volume * density

mass of water = (60.0 mL) * (1.00 g/mL)

mass of water = 60.0 g

Initial temperature = 11.0 ^oC

The shortcut method to find the final temperature is

final temperature = [(M₁ * T₁) + (M₂ * T₂)] / (M₁ + M₂)

where M₁ = mass of water in cup #1 = 80.0 g

T₁ = initial temperature of water in cup #1 = 96.0 ^oC

M₂ = mass of water in cup #2 = 60.0 g

T₂ = initial temperature of water in cup #2 = 11.0 ^oC

final temperature = [(80.0 g * 96.0 ^oC) + (60.0 g * 11.0 ^oC)] / (80.0 g + 60.0 g)

final temperature = (8340 g.^oC) / (140.0 g)

final temperature = 59.6 ^oC

This method is only valid when liquid in cup #1 and cup #2 is same