Assume you are performing the calibration step and you begin with 70 g of water at 20*C and 70 g of water at 80*C. 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 to be 45*C. What is the heat capacity of the calorimeter?
To determine the heat capacity of the calorimeter, we can use the principle of heat exchange:
Q₁ (heat gained) = Q₂ (heat lost)
The heat gained by the water at 20°C can be calculated using the formula:
Q₁ = mcΔT
where:
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
The heat lost by the water at 80°C can be calculated in the same way.
Since no heat is lost or gained by the calorimeter itself, we can write:
Q₁ + Q₂ = 0
This means that the heat gained by the water at 20°C is equal to the heat lost by the water at 80°C.
Let's calculate the heat gained and lost:
First, for the water at 20°C:
m₁ = 70 g (mass of water at 20°C)
c₁ = 4.186 J/g°C (specific heat capacity of water)
ΔT₁ = (45°C - 20°C) = 25°C (change in temperature)
Q₁ = m₁c₁ΔT₁
= 70 g * 4.186 J/g°C * 25°C
= 69,255 J
Next, for the water at 80°C:
m₂ = 70 g (mass of water at 80°C)
c₂ = 4.186 J/g°C (specific heat capacity of water)
ΔT₂ = (80°C - 45°C) = 35°C (change in temperature)
Q₂ = m₂c₂ΔT₂
= 70 g * 4.186 J/g°C * 35°C
= 102,879 J
Since Q₁ + Q₂ = 0, we can write:
Q₁ + (-Q₁) = 0
69,255 J + (-102,879 J) = 0
From this equation, we can conclude that the heat gained is equal to the heat lost. Therefore, the heat capacity of the calorimeter is 102,879 J.