How do you do this?
If 100 grams of boiling water at 100 degrees C is poured into a 100 g beaker of platinum at 0 degrees C, an intermediate temperature will result. Knowing that the specific heat of water is about 32 times larger than the specific heat of platinum, approximate the resulting temperate?
Heat lost by water + heat gained by Pt = 0
We call Pt 1; then H2O = 32
[mass H2O x 32 x (Tfinal-Tinitial)] + [mass Pt x 1 x (Tfinal-Tinitial) = 0
Substitute and solve for Tfinal.
DrBob222: Can you check my recent lab post. I'm just confused on what numbers to put where because some things aren't given. Lab due tmrw. Please help! Thankyou so much.
Well, that certainly sounds like a puzzling situation! Let me see if I can clown around with the numbers for you.
Since the specific heat of water is approximately 32 times larger than that of platinum, we can assume that the water will have a much greater influence on the final temperature.
Without getting too technical, let's agree that the water is the life of the party here! So, imagine the water jumping into the beaker, waving its hands in excitement. It's like a wild water slide party!
Given that, I would approximate the resulting temperature to be somewhere between the initial temperatures of the water and platinum. Not too hot, not too cold, but just right for a temperature that both elements can agree on.
Now, if I were to put my clown goggles on, I'd say this party temperature would be around 50 degrees C. But remember, this is just a rough approximation - not exactly a scientific answer. So, take it with a pinch of clown dust!
Keep in mind that this is just a playful answer and not something you can take too seriously. For a more accurate calculation, you might need the assistance of a qualified scientist. Happy clowning!
To approximate the resulting temperature, you can use the principle of conservation of energy, which states that the total energy before and after the transfer is the same.
Here are the steps:
1. Calculate the amount of heat gained by the water:
- The specific heat capacity of water is approximately 4.18 J/g°C, and the initial temperature of the water is 100°C.
- The mass of the water is 100 grams.
- So, the heat gained by the water is: Qwater = (4.18 J/g°C) * 100 g * (Tfinal - 100°C)
(Note: Tfinal is the final temperature we are trying to approximate.)
2. Calculate the amount of heat lost by the platinum:
- The specific heat capacity of platinum is approximately 0.13 J/g°C, and the initial temperature of the platinum is 0°C.
- The mass of the platinum is 100 grams.
- So, the heat lost by the platinum is: Qplatinum = (0.13 J/g°C) * 100 g * (Tfinal - 0°C)
3. Apply the conservation of energy principle:
- According to the principle of conservation of energy, the heat gained by the water is equal to the heat lost by the platinum:
Qwater = Qplatinum
(4.18 J/g°C) * 100 g * (Tfinal - 100°C) = (0.13 J/g°C) * 100 g * (Tfinal - 0°C)
4. Solve for Tfinal:
- Distribute and simplify the equation: 418 (Tfinal - 100) = 13 (Tfinal - 0)
- Solve for Tfinal:
418Tfinal - 41800 = 13Tfinal
405Tfinal = 41800
Tfinal ≈ 103.21°C
Therefore, the approximate resulting temperature is 103.21°C.
To approximate the resulting temperature when 100 grams of boiling water at 100 degrees Celsius is poured into a 100 g beaker of platinum at 0 degrees Celsius, we can use the principle of heat transfer and the concept of specific heat.
1. First, we need to understand the principle of heat transfer. When two objects with different temperatures come into contact, heat will flow from the object at a higher temperature to the object at a lower temperature until they reach thermal equilibrium (the same temperature).
2. Next, we need to consider the specific heat capacities of water and platinum. Specific heat is the amount of heat energy required to raise the temperature of a given substance by a certain amount. Given that the specific heat capacity of water is about 32 times larger than that of platinum, we can assume that it takes much more heat energy to raise the temperature of water compared to platinum.
Now, let's calculate the approximate resulting temperature using the principles mentioned above:
1. Calculate the heat transfer from the boiling water:
The heat transferred (Q) can be calculated using the formula:
Q = m * c * ΔT
Where:
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature of the substance (in °C)
In this case, the mass of the boiling water (m1) is 100 grams, the specific heat capacity of water (c1) is approximately 4.18 J/g°C, and the change in temperature (ΔT1) is 100°C - 0°C = 100°C.
Q1 = m1 * c1 * ΔT1
2. Calculate the heat transfer to the platinum beaker:
Using the same formula, we can calculate the heat transferred (Q2) to the platinum beaker. The mass of the platinum beaker (m2) is also 100 grams, and the specific heat capacity of platinum (c2) is approximately 0.13 J/g°C. The change in temperature (ΔT2) for the platinum beaker is the temperature that we want to find.
Q2 = m2 * c2 * ΔT2
3. At thermal equilibrium, the total heat transferred by the boiling water (Q1) will be equal to the total heat transferred to the platinum beaker (Q2). Therefore, we can set up the equation:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
4. Substitute the known values into the equation:
Substituting the masses, specific heat capacities, and change in temperatures into the equation, we have:
100 g * 4.18 J/g°C * 100°C = 100 g * 0.13 J/g°C * ΔT2
Simplify the equation:
418 J/°C = 13 J/°C * ΔT2
5. Solve for ΔT2:
To find ΔT2, we need to isolate it on one side of the equation. Divide both sides of the equation by 13 J/°C:
ΔT2 = 418 J/°C / 13 J/°C
ΔT2 ≈ 32.15°C
Therefore, the approximate resulting temperature after pouring boiling water into the platinum beaker would be approximately 32.15°C.