dissolving 4.022 g of a salt in 100 mL of water changes temperature by -1.2^oC: what is the heat absorbed?

The specific heat of water is 4.184 J/(g*C)

To find the heat absorbed when dissolving 4.022 g of a salt in 100 mL of water, you can follow these steps:

1. Convert the volume of water from milliliters (mL) to grams (g) by using the density of water, which is 1 g/mL. So, 100 mL of water is equivalent to 100 g.

2. Calculate the temperature change (ΔT) by subtracting the initial temperature from the final temperature. In this case, the temperature change is -1.2°C.

3. Use the equation Q = m * ΔT * C to calculate the heat absorbed, where Q is the heat absorbed, m is the mass of the water, ΔT is the temperature change, and C is the specific heat of water.

4. Substitute the given values into the equation:

Q = (100 g) * (-1.2°C) * (4.184 J/(g*C))

5. Multiply the values inside the parentheses:

Q = (-120.0 J/g°C) * (4.184 J/(g°C))

6. Calculate the final value:

Q ≈ -501.48 J

Therefore, the heat absorbed when dissolving 4.022 g of a salt in 100 mL of water is approximately -501.48 Joules.