What is the final temperature and physical state of water when 250 grams of water at 85 degrees C is added to 80.0 grams of ice at -15 degree C?
Is the physical state of water liquid?
I don't understand the set up that I have to set the equation equal to 0 ...
What is that set up. If can see what you're talking about I can walk you through each step.
To determine the final temperature and physical state of water in this scenario, we need to apply the principle of conservation of energy and use the equation:
mcΔT + mcΔT = 0
where:
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
In this case, we have two substances: water and ice.
For the water at 85 degrees Celsius:
m1 = 250 grams
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = Tf - 85 (Tf is the final temperature of water)
For the ice at -15 degrees Celsius:
m2 = 80.0 grams
c2 = 2.09 J/g°C (specific heat capacity of ice)
ΔT2 = Tf - (-15)
Substituting these values into the equation:
(250 grams)(4.18 J/g°C)(Tf - 85) + (80.0 grams)(2.09 J/g°C)(Tf - (-15)) = 0
To solve this equation, we can simplify it:
(1045(Tf - 85)) + (167.2(Tf + 15)) = 0
1045Tf - 88625 + 167.2Tf + 2514 = 0
1212.2Tf - 86111 = 0
1212.2Tf = 86111
Tf = 86111 / 1212.2
Tf ≈ 71.03°C
Hence, the final temperature of water is approximately 71.03 degrees Celsius.
Regarding the physical state of water, we need to compare the final temperature (71.03°C) with the boiling point (100°C) and melting point (0°C) of water. Since the final temperature is below 100°C, we can conclude that the physical state of water is in liquid form.
Therefore, the final temperature is approximately 71.03 degrees Celsius, and the physical state of water is liquid.