Question

A large punch bowl holds 3.20 kg of lemonade (which is essentially water) at 24.0 ∘C. A 5.40×10^−2-kg ice cube at -10.0 ∘C is placed in the lemonade. What is the final temperature of the system?

Answer 1

The final temperature of the system, after adding the ice cube to the lemonade, is approximately 24.04 ∘C.

Step-by-step explanation:

To find the final temperature of the system, we can use the equation m1c1ΔT1 = m2c2ΔT2, where m1 and m2 are the masses of the substances, c1 and c2 are their specific heat capacities, and ΔT1 and ΔT2 are the changes in temperature.

In this case, the mass of the lemonade is 3.20 kg, the mass of the ice cube is 5.40×10^−2 kg, the initial temperature of the lemonade is 24.0 ∘C, and the initial temperature of the ice cube is -10.0 ∘C.

Let's assume the final temperature of the system is T.

Using the equation, we can write:

(3.20 kg)(c1)(T - 24.0 ∘C) = (5.40×10^−2 kg)(c2)(T - (-10.0 ∘C))

Now, we need to know the specific heat capacities of water and ice. The specific heat capacity of water is approximately 4.18 J/g⋅°C, and the specific heat capacity of ice is approximately 2.09 J/g⋅°C.

Converting the masses to grams, we have:

(3200 g)(4.18 J/g⋅°C)(T - 24.0 ∘C) = (0.054 g)(2.09 J/g⋅°C)(T - (-10.0 ∘C))

Simplifying the equation, we get:

13376(T - 24.0) = 0.11286(T + 10.0)

Expanding and rearranging the equation, we have:

13376T - 321024 = 0.11286T + 1.1286

Combining like terms, we get:

13376T - 0.11286T = 321024 + 1.1286

Simplifying further, we have:

13375.88714T = 321025.1286

Dividing both sides by 13375.88714, we get:

T ≈ 24.04 ∘C

Therefore, the final temperature of the system is approximately 24.04 ∘C.