Assume you begin with 60 g of water at 20 oC and 60 g of water at 80 oC. After adding the two portions of water into your calorimeter, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter?

To determine the heat capacity of the calorimeter, we need to use the principle of heat transfer, which states that the heat gained by one substance is equal to the heat lost by another substance.

In this scenario, we have two portions of water at different temperatures that are mixed in the calorimeter. The heat lost by the hot water is equal to the heat gained by the cold water and the calorimeter.

First, let's calculate the heat lost by the hot water. We use the formula:

Q = m * c * ΔT

Where:
Q is the heat lost or gained
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

For the hot water:
m = 60 g (given)
c = specific heat capacity of water (approximately 4.18 J/g°C)
ΔT = (80°C - 45°C) = 35°C

Q_hot = 60 g * 4.18 J/g°C * 35°C
Q_hot = 8778 J

Since heat is conserved, the heat gained by the cold water and the calorimeter is also 8778 J.

Now, let's calculate the heat gained by the cold water. Again, using the formula:

Q_cold = m * c * ΔT

For the cold water:
m = 60 g (given)
c = specific heat capacity of water (approximately 4.18 J/g°C)
ΔT = (45°C - 20°C) = 25°C

Q_cold = 60 g * 4.18 J/g°C * 25°C
Q_cold = 6270 J

Finally, to determine the heat gained by the calorimeter, we subtract the heat gained by the cold water from the total heat transferred:

Q_calorimeter = Q_total - Q_cold
Q_calorimeter = 8778 J - 6270 J
Q_calorimeter = 2508 J

Therefore, the heat capacity of the calorimeter is 2508 J.

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