Question

In a calorimeter, 0.300 kg of water at 80.0∘ C is added to 1.00 kg of water at 20.0∘C. What will be the final temperature of the mixture when it reaches thermal equilibrium?

Select one:
a. 63.8∘ C
b. 40.8 ∘ C
c. 33.8 ∘ C
d. 21.8 ∘ C
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Answer 1

In a calorimeter, 0.300 kg of water at 80.0∘ C is added to 1.00 kg of water at 20.0∘C. In this case, the final temperature is approximately C. 33.8∘C.

Step-by-step explanation:

In a calorimeter, the final temperature of a mixture can be determined using the principle of thermal equilibrium.

In this problem, we have 0.300 kg of water at 80.0∘C and 1.00 kg of water at 20.0∘C.

The heat lost by the hot water is equal to the heat gained by the cold water at thermal equilibrium.

We can use the equation Q = mcΔT to calculate the heat lost and gained,

where

Q is the heat, m is the mass

c is the specific heat capacity

ΔT is the change in temperature.

Let's calculate the heat lost by the hot water:

Qhot = mcΔThot

Qhot = (0.300 kg)(4186 J/kg⋅K)(80.0∘C - T)

Similarly, the heat gained by the cold water is:

Qcold = mcΔTcold

Qcold = (1.00 kg)(4186 J/kg⋅K)(T - 20.0∘C)

At thermal equilibrium, Qhot = Qcold. Setting these two equations equal to each other, we can solve for T:

(0.300 kg)(4186 J/kg⋅K)(80.0∘C - T) = (1.00 kg)(4186 J/kg⋅K)(T - 20.0∘C)

After solving this equation, we find that T ≈ 33.8∘C. Therefore, the correct answer is option C. 33.8∘C.