Assume you are performing the calibration step of Experiment 8 and you begin with 70 g of water at 20 oC and 70 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter
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Writeacher is correct. I never would have clicked on uah.
How much heat was gained by the cold water?
mass x specific heat x (Tfinal-Tinitial) = ??
How much was lost by the hot water?
mass x specific heat x (Tfinal-Tinitial) = ??
What's the difference? The difference is how much was absorbed by the calorimeter. Take the difference and divide by delta T to find cCal. You don't say what the initial T of the calorimeter was; perhaps it was 20, the same as the cold water. Perhaps it was something different but delta T will be Tfinal (45 C?) - Tinitial.
Well, looks like we've got a hot mess here, or rather a cool one in this case! Let's see if we can bring some order to the chaos.
So, we start with 70 g of water at 20°C and 70 g of water at 80°C. When we mix these two together, we end up with a lukewarm temperature of 45°C. Clearly, these waters didn't hit it off quite as well as we hoped.
Now, we want to find the heat capacity of the calorimeter, or in simpler terms, how much heat this thing can handle before it gets too hot to handle. It's like finding the tolerance level of a person trying spicy food for the first time. Some can handle the heat, while others might literally lose their cool.
To find the heat capacity, we need to apply a little science. We know that the heat lost by the hot water is equal to the heat gained by the cold water. So, we can use the equation:
(mass of cold water) * (specific heat capacity of water) * (change in temperature of cold water) = (mass of hot water) * (specific heat capacity of water) * (change in temperature of hot water)
Plugging in the numbers, we get:
(70 g) * (specific heat capacity of water) * (45°C - 20°C) = (70 g) * (specific heat capacity of water) * (80°C - 45°C)
Before we proceed, let's pause for a moment and appreciate the fact that water can be moody too, with its temperature swinging from one extreme to another. It's probably heating things up or cooling things down just for attention.
Anyway, let's solve this equation and find the specific heat capacity of the calorimeter. And with a sprinkle of math and a dash of humor, the answer might just pop out of the hat. Voila!
To find the heat capacity of the calorimeter, we need to use the principle of heat transfer and the equation Q = mcΔT, where Q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have two portions of water with different temperatures that are mixed in the calorimeter. The heat transfer between the two portions can be calculated using the equation:
Q = mcΔT
First, let's calculate the heat transfer from the hot water to the mixed water. The mass of hot water is 70 g, the specific heat capacity of water is 4.18 J/g°C (or 4.18 J/gK), and the change in temperature is (80 - 45) = 35°C. So, for the heat transfer from the hot water, we have:
Q1 = (70 g) × (4.18 J/g°C) × (35°C)
Next, let's calculate the heat transfer from the cold water to the mixed water. The mass of the cold water is 70 g, the specific heat capacity of water is 4.18 J/g°C (or 4.18 J/gK), and the change in temperature is (45 - 20) = 25°C. So, for the heat transfer from the cold water, we have:
Q2 = (70 g) × (4.18 J/g°C) × (25°C)
Now, since the heat lost by the hot water is equal to the heat gained by the cold water, we can set up an equation:
Q1 = Q2
(70 g) × (4.18 J/g°C) × (35°C) = (70 g) × (4.18 J/g°C) × (25°C)
Now, we can solve this equation to find the heat capacity of the calorimeter:
(35°C) = (25°C)
(35°C) = (25°C) / (70 g) × (4.18 J/g°C)
On simplifying, we find:
(35°C) = C × (70 g) × (4.18 J/g°C)
Here, C represents the heat capacity of the calorimeter, which is the value we need to calculate. By rearranging this equation, we can solve for C:
C = (35°C) / [(70 g) × (4.18 J/g°C)]
On evaluating this expression, we can find the heat capacity of the calorimeter.