A 10 gram sample of water is heated to 105 ℃ and is mixed with a 25 gram sample of water cooled to 25℃. What is the final temperature of the water mixture?

The final temperature of the water mixture is 47.85°C

Explanation:

Given,

For Warm Water

mass = 10grams

Temperature = 105°C

For Cold Water

mass = 25grams

Temperature = 25°C

When a sample of warm water is mixed with a sample of cool water,

The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

Cp = Specific heat of water = 4.184 J/Kg°C

So,

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using is FINAL temperature. This is what we are solving for.

The warmer water goes down from to 105°C to x, so this means its Δt equals 105°C - x. The colder water goes up in temperature, so its Δt equals x - 25℃

Substituting the values,

(10) (105°C - x) (4.184) = (25) (x - 25℃) (4.184)

Solving for x, we get

x = 47.85°C

Therefore, The final temperature of the water mixture is 47.85°C.