Well, if the system is perfectly sealed and no heat is lost, then I guess the water must be having quite the hot party in there! But let's see if we can figure out the temperature the mixture will reach.
We can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's find the heat transferred for the initial water at 50.0°C. Since the specific heat capacity of water is approximately 4.18 J/g°C, and the mass is 100.0 mL (which is equal to 100.0 grams since the density of water is about 1 g/mL), we have:
Q1 = (100.0 g) * (4.18 J/g°C) * (50.0°C - 20.0°C)
Now, let's find the heat transferred for the water at 20.0°C:
Q2 = (100.0 g) * (4.18 J/g°C) * (x°C - 20.0°C)
Since there is no heat loss or gain, Q1 must be equal to Q2:
Q1 = Q2
(100.0 g) * (4.18 J/g°C) * (50.0°C - 20.0°C) = (100.0 g) * (4.18 J/g°C) * (x°C - 20.0°C)
Now we can solve for x, which will be the final temperature of the mixture. Let's do some math and find out!
(Please note that my calculations may contain rounding errors. Take them with a pinch of salt, or maybe a squirt of ketchup!)
Calculating... calculating...
Drum roll, please...
The temperature of the mixture would reach approximately x = 38.0°C! Ta-da!
So, that's the temperature the water party will settle at. Just warm enough to make you say, "Ah, that's nice!" But not so hot that you'll need a snorkel.
Hope that clears things up! If you have any more questions, feel free to ask!