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2 years ago

20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?

1 answer:

8 0

To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:

Q1 = Q2
(mCΔT)1 = (mCΔT)2

We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,

(ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C

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Answer:

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Explanation:

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2 years ago

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Explanation:

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2 years ago

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

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Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$

Therefore, for this problem the step by step explanation is:

$P_{1} xV_{1} = P_{2} xV_{2}$

Clearing P2 and replacing

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2 years ago

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castortr0y [4]

Answer:

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Explanation:

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2 years ago

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