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 Tf= 109 C?
Is it in the steam phase?
How could it be in the steam phase? Is the cold ice going to make it boil? And, it is not 109, it cant heat up.
The sum of heats gained =0
80(cice)(0+15)+80Hf+80cw(Tf-0)+250cw(Tf-85)=0
solve for Tfinal Tf
To determine the final temperature and physical state of the water when it is mixed with the ice, we can use the principles of heat transfer and the concept of latent heat.
First, let's calculate the heat absorbed or released during the phase changes and temperature change for both the water and the ice.
1. For the water at 85 degrees C:
The heat absorbed (q) can be calculated using the formula:
q = m * c * ΔT,
where
m = mass of the water (250 grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
q = 250 * 4.18 * (Tf - 85)
2. For the ice at -15 degrees C:
The heat released (q) due to phase change can be calculated using the formula:
q = m * ΔHf,
where
m = mass of the ice (80.0 grams)
ΔHf = heat of fusion of water (334 J/g)
q = 80.0 * 334
3. For the ice melting and heating up to the final temperature:
The heat absorbed (q) can be calculated using the formula:
q = m * c * ΔT,
where
m = mass of the water (80.0 grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - 0)
q = 80.0 * 4.18 * Tf
Now, let's equate the heat absorbed and the heat released:
250 * 4.18 * (Tf - 85) = (80.0 * 334) + (80.0 * 4.18 * Tf)
Solving this equation will give us the final temperature (Tf).
After getting the final temperature, we can determine the physical state of water at that temperature.
As an AI bot, I can assist you in solving the equation and providing you with the final temperature and physical state of water. However, the specific calculations will depend on the values of the constants mentioned above.