What is the final temperature upon mixing 75.0 grams of water held at 45.0 ° C witH₂ 5.0 grams of water held at 90.0 ° C?

Question

Quizplus · Accepted Answer

To solve this problem, we can use the formula:

Q = m * c * ΔT

where Q is the heat absorbed/released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the heat absorbed by the colder water:

Q1 = 75.0 g * 4.18 J/g°C * (Tf - 45.0°C)

where Tf is the final temperature.

Next, let's calculate the heat released by the hotter water:

Q2 = 5.0 g * 4.18 J/g°C * (90.0°C - Tf)

Note that the temperature difference is 90.0°C - Tf because the initial temperature of the hotter water is greater than Tf.

Since energy is conserved in the process, Q1 = Q2.

75.0 g * 4.18 J/g°C * (Tf - 45.0°C) = 5.0 g * 4.18 J/g°C * (90.0°C - Tf)

3135 Tf - 141075 = 1674 - 20.9 Tf

Simplifying and solving for Tf gives:

34.3 Tf = 142749

Tf ≈ 56.3°C

Therefore, the final temperature is 56.3°C, which is choice C.

What Is the Final Temperature Upon Mixing 75