In a constant-pressure calorimetry experiment, a reaction gives off 20.8 kJ of heat. The calorimeter contains 150 g of water, initially at 21.4 degrees Celsius. What is the final temperature of the water? The heat capacity of the calorimeter is negligibly small.
20,800 J = mass H2O x specific heat H2O x (Tfinal-Tinitial)
To find the final temperature of the water in the calorimeter, we can use the concept of heat transfer and the equation of heat transfer, q = mcΔT, where:
- q is the heat transferred
- m is the mass of the substance (water in this case)
- c is the specific heat capacity of the substance (water)
- ΔT is the change in temperature
In this case, we want to find the final temperature, so we rearrange the equation to solve for ΔT:
ΔT = q / (mc)
First, let's find the heat transferred (q) using the given information: q = 20.8 kJ.
Next, let's find the mass of water (m), which is given as 150 g.
The specific heat capacity of water (c) is approximately 4.18 J/g°C.
Now, we can substitute the values into the equation to find the change in temperature (ΔT):
ΔT = (20.8 kJ) / (150 g * 4.18 J/g°C)
To simplify the calculation, let's convert 20.8 kJ to J:
20.8 kJ = 20.8 × 10^3 J
Substituting the values and calculating:
ΔT = (20.8 × 10^3 J) / (150 g * 4.18 J/g°C)
ΔT ≈ 31.36 / (150 * 4.18)
ΔT ≈ 31.36 / 627
ΔT ≈ 0.05°C
Since the initial temperature is 21.4°C, the final temperature is approximately:
Final temperature = Initial temperature + ΔT
Final temperature = 21.4°C + 0.05°C
Final temperature ≈ 21.45°C
Therefore, the final temperature of the water in the calorimeter is approximately 21.45 degrees Celsius.