223k views
5 votes
Calculate AE and AH when the temperature of one mole of water is increased from 10°C to 70°C. The density of water is 0.9778 g cm³ and 0.9997 g cm³ at 70°C and 10°C respectively.​

User Smhx
by
8.2k points

1 Answer

4 votes

Answer:

To calculate AE and AH when the temperature of water is increased from 10°C to 70°C, we can use the formula:

AE = m × c × ΔT

Where:

- AE is the change in internal energy

- m is the mass of water

- c is the specific heat capacity of water

- ΔT is the change in temperature

First, we need to calculate the mass of water. We know that the density of water is 0.9997 g/cm³ at 10°C and 0.9778 g/cm³ at 70°C. Since the density is mass divided by volume, we can rearrange the formula to solve for mass:

mass = density × volume

We can assume the volume of water remains constant. Therefore, the mass at 10°C is:

mass₁ = density₁ × volume

And the mass at 70°C is:

mass₂ = density₂ × volume

Now, we need to calculate the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

Using the given values, we can calculate the change in internal energy:

AE = (mass₂ × c × (70°C - 10°C)) - (mass₁ × c × (70°C - 10°C))

Finally, to calculate AH (the enthalpy change), we need to consider any phase changes that occur. Water undergoes a phase change from liquid to gas at 100°C. If the final temperature is below 100°C, there is no phase change and AH is equal to AE. However, if the final temperature is above 100°C, we need to consider the energy required for the phase change.

In this case, since the final temperature is 70°C, AH is equal to AE.

User ODelibalta
by
7.9k points