If 25.0 g H₂O at 85.0°C is mixed in a coffee-cup calorimeter with 15.0 g H₂O at 20.0°C,what is the final temperature of the mixture? The specific heat of water is 4.18 J/g⋅°C.

Question

Quizplus · Accepted Answer

1. First, calculate the heat absorbed by the colder water:

q = mCΔT = 15.0 g x 4.18 J/g⋅°C x (Tf - 20.0°C)

2. Next, calculate the heat released by the warmer water:

q = mCΔT = 25.0 g x 4.18 J/g⋅°C x (85.0°C - Tf)

Note that both equations use the same specific heat capacity of water (4.18 J/g⋅°C).

3. Since the heat released and absorbed must be equal (assuming no heat is exchanged with the surroundings), we can set the two equations equal to each other and solve for Tf:

15.0 g x 4.18 J/g⋅°C x (Tf - 20.0°C) = 25.0 g x 4.18 J/g⋅°C x (85.0°C - Tf)

62.70Tf - 1254 = 4391.50 - 104.50Tf

167.20Tf = 5645.50

Tf = 33.72°C

Therefore, the final temperature of the mixture is:

Tf + 20.0°C = 53.72°C, which is closest to choice C (60.6°C).

If 250 G H2O at 85

If 250 G H₂O at 85