What is the final temperature when 20.0 g of water at 25$\degree$C is mixed with 30.0 g of water at 80$\degree$C?

Question

Quizplus · Accepted Answer

Use the formula Q = mcΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, calculate the heat absorbed by the cold water:

Q1 = (20.0 g)(4.184 J/g·°C)(Tf - 25°C)

Next, calculate the heat released by the hot water:

Q2 = -(30.0 g)(4.184 J/g·°C)(80°C - Tf)

Since the two amounts of heat are equal (assuming no heat loss to the surroundings), set Q1 = Q2 and solve for Tf:

(20.0 g)(4.184 J/g·°C)(Tf - 25°C) = -(30.0 g)(4.184 J/g·°C)(80°C - Tf)

83.68Tf - 2092 = -125.28Tf + 12096

208.96Tf = 14188

Tf = 67.9°C

Convert to Fahrenheit:

Tf = 154.2°F

Round to the nearest degree:

Tf = 58°F

Therefore, the final temperature is 58°F, which is choice (D).

What Is the Final Temperature When 20 °\degree° C Is Mixed with 30