A piece of copper of mass 21.3 g at 106.0 °C is placed in a vessel of negligible heat capacity but containing 53.0 g of water at 19.3 °C. Calculate the final temperature of the water.

I keep getting 22 celsius and I'm not sure where I'm going wrong. I plug it into the equation grams*specific heat capacity*change in t.

If you had typed in what you did I could have found the error by now. I don't know what you did wrong either. How do you know 22 is wrong? What are you using for specific heat Cu.

To solve this question, you need to use the principle of energy conservation in a calorimetry experiment.

The equation you referred to, grams * specific heat capacity * change in temperature, is the correct formula to use. However, you also need to account for the heat gained by the water (m1c1ΔT1) and the heat lost by the copper (m2c2ΔT2), where m1 and m2 are the masses of water and copper, c1 and c2 are their respective specific heat capacities, and ΔT1 and ΔT2 are the changes in temperature.

Let's break down the steps:

1. Calculate the heat gained by the water:
Q1 = m1 * c1 * ΔT1
Here, m1 = 53.0 g (mass of water), c1 = 4.18 J/g°C (specific heat capacity of water), and ΔT1 is the change in temperature of the water, which is Tf - 19.3 °C (final temperature minus initial temperature).
So, Q1 = 53.0 g * 4.18 J/g°C * (Tf - 19.3 °C)

2. Calculate the heat lost by the copper:
Q2 = m2 * c2 * ΔT2
Here, m2 = 21.3 g (mass of copper), c2 = 0.39 J/g°C (specific heat capacity of copper), and ΔT2 is the change in temperature of the copper, which is Tf - 106.0 °C (final temperature minus initial temperature).
So, Q2 = 21.3 g * 0.39 J/g°C * (Tf - 106.0 °C)

3. Since energy is conserved, the heat gained by the water is equal to the heat lost by the copper:
Q1 = Q2
We can equate the two equations obtained above and solve for Tf.

Now, by substituting the given values into the equation and solving for Tf, you will get the final temperature of the water.