Assume you are performing the calibration step of Experiment 8 and you begin with 50 g of water at 20 oC and 50 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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To determine the heat capacity of the calorimeter, we can use the principle of energy conservation. The heat lost by the hot water will be equal to the heat gained by the cold water and the calorimeter.
First, we need to calculate the heat gained by the cold water. We can use the equation:
q = m * c * ΔT
where q is the heat gained (or lost), m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
In this case, the cold water gains heat, so:
q_cold = m_cold * c_water * ΔT_cold
We are given that the cold water starts at 20 oC, and the final temperature after mixing is 45 oC. We also know the mass of the cold water is 50 g.
ΔT_cold = 45 oC - 20 oC = 25 oC
Now, we need to calculate the heat lost by the hot water. Similarly:
q_hot = m_hot * c_water * ΔT_hot
The hot water starts at 80 oC and ends up at the same final temperature of 45 oC. The mass of the hot water is also 50 g.
ΔT_hot = 80 oC - 45 oC = 35 oC
Since the heat lost by the hot water must be equal to the heat gained by the cold water, we have:
q_hot = q_cold
m_hot * c_water * ΔT_hot = m_cold * c_water * ΔT_cold
The mass of the water cancels out:
m_hot * ΔT_hot = m_cold * ΔT_cold
Now, we can rearrange the equation to solve for the specific heat capacity of the calorimeter, c_calorimeter:
c_calorimeter = (m_cold * c_water * ΔT_cold) / (m_hot * ΔT_hot)
Plugging in the given values, we get:
c_calorimeter = (50 g * 4.18 J/g°C * 25 oC) / (50 g * 35 oC)
Calculating this gives us the heat capacity of the calorimeter.