i need help setting this problem up.
You have an insulated container of water that has a temperature of 83.2 C, you continue to add ice until the temperature reaches .6 C and you remove the excess ice, your volume increased by 79.8 mL. What amount of hot water did you have in the container to start with?
If we assume the ice is at zero degrees C (which may or may not be so) AND that 79.8 g ice melts to produce 79.8 g water (at zero C, which isn't so because the density of ice is not 1.00 g/mL. Then
massH2O x specificheatwater x (Tf - Ti) + waterfromice x specificheat x (Tf - Ti) + massice to melt x heat fusion ice = 0
massH2O x 4.184 x (83.2-6.0) + 79.8 x 4.184 x (6.0-0.0) + 79.8 x 334 = 0
Solve for massH2O. Check my thinking. Check the numbers.
Ignore thermal expansion.
Assume the volume increase is due to melted ice. I don't think this problem can be done unless an initial temperature of the ice is known.
To solve this problem, you need to use the principle of conservation of energy. Here's how you can set up the problem:
1. Identify the variables:
- massH2O: the mass of hot water in the container to start with (what we need to find)
- specificheatwater: the specific heat capacity of water (typically 4.184 J/g·°C)
- Tf: the final temperature of the system after adding ice (0.6 °C)
- Ti: the initial temperature of the system (83.2 °C)
- waterfromice: the mass of water that results from the melted ice (79.8 g)
- specificheat: the specific heat capacity of ice (also 4.184 J/g·°C, assuming the ice is at 0 °C)
- massice to melt: the mass of ice that melts to produce the water (also 79.8 g)
2. Apply the principle of conservation of energy:
The total energy change in the system is equal to zero. This can be represented by the equation:
massH2O x specificheatwater x (Tf - Ti) + waterfromice x specificheat x (Tf - Ti) + massice to melt x heat fusion ice = 0
3. Substitute the known values into the equation:
massH2O x 4.184 x (0.6 - 83.2) + 79.8 x 4.184 x (0.6 - 0.0) + 79.8 x 334 = 0
4. Solve for massH2O:
Rearrange the equation and solve for massH2O:
massH2O x 4.184 x (-82.6) = -79.8 x 4.184 x (0.6) - 79.8 x 334
massH2O = (-79.8 x 4.184 x (0.6) - 79.8 x 334) / (4.184 x (-82.6))
5. Calculate the massH2O:
Plug in the values and solve for massH2O:
massH2O ≈ 1005 grams (rounded to three significant figures)
Therefore, you had approximately 1005 grams of hot water in the insulated container to start with.