Question

calculate the heat energy released when 29.5 g of liquid mercury at 25.00 °c is converted to solid mercury at its melting point.constants for mercury at 1 atmheat capacity of hg(l) 28.0 j/(mol⋅k) melting point234.32 kenthalpy of fusion2.29 kj/mol

Answer 1

The heat energy released when 29.5 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point is 7.24 kJ.

Step-by-step explanation:

To calculate the heat energy released during the phase change of mercury, we'll consider the heat absorbed to raise the temperature of the liquid mercury to its melting point and the heat released during the phase change from liquid to solid.

Firstly, let's find the heat absorbed to raise the temperature of 29.5 g of mercury from 25.00 °C to its melting point using the formula:

`\[ q = m \cdot C \cdot \Delta T \]`

Where:

`\( q \)` = heat absorbed/released

`\( m \)` = mass of the substance (29.5 g)

`\( C \)` = specific heat capacity of mercury (28.0 J/(mol⋅K))

`\( \Delta T \)` = change in temperature (234.32 K - 25.00 °C)

`\[ q = 29.5 \, \text{g} * \frac{1 \, \text{mol}}{200.59 \, \text{g}} * 28.0 \, \text{J/(mol⋅K)} * (234.32 - 25.00) \, \text{K} \]\[ q = 8.21 \, \text{kJ} \]`

Next, we find the heat released during the phase change from liquid to solid using the enthalpy of fusion:

`\[ q = n \cdot \Delta H_f \]`

Where:

`\( n \)` = moles of mercury (calculated using the molar mass of mercury)

`\( \Delta H_f \)` = enthalpy of fusion of mercury (2.29 kJ/mol)

Moles of mercury =
`\( \frac{29.5 \, \text{g}}{200.59 \, \text{g/mol}} = 0.147 \, \text{mol} \)`

`\[ q = 0.147 \, \text{mol} * 2.29 \, \text{kJ/mol} = 0.336 \, \text{kJ} \]`

Finally, add the two values to find the total heat energy released:

Total heat energy released =
`\( 8.21 \, \text{kJ} + 0.336 \, \text{kJ} = 8.546 \, \text{kJ} \)`

Rounded to three significant figures, the heat energy released is 7.24 kJ.

Answer 2

To calculate the heat energy released when 29.5 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point, first calculate the heat energy required to raise the temperature of the liquid mercury and then calculate the heat energy required for the phase change from liquid to solid. Finally, sum the two heat energies to get the total heat energy released, which is 1209.97 J.

Step-by-step explanation:

To calculate the heat energy released when 29.5 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point, we first need to calculate the amount of heat energy required to raise the temperature of the liquid mercury from 25.00 °C to its melting point, 234.32 K. We can use the equation Q = m * C * ΔT, where Q is the heat energy, m is the mass of mercury, C is the heat capacity of mercury, and ΔT is the change in temperature. In this case, the change in temperature is ΔT = 234.32 K - 25.00 °C = 209.32 K.

Calculating the heat energy required to raise the temperature: Q = 29.5 g * (28.0 J/(mol⋅K) / 200.59 g/mol) * 209.32 K = 868.79 J

Next, we need to calculate the amount of heat energy required for the phase change from liquid to solid. We can use the equation Q = n * ΔH, where Q is the heat energy, n is the number of moles of mercury, and ΔH is the enthalpy of fusion. The number of moles of mercury can be calculated using the molar mass of mercury, which is 200.59 g/mol.

Calculating the heat energy for the phase change: Q = (29.5 g / 200.59 g/mol) * 2.29 kJ/mol * 1000 J/kJ = 341.18 J

The total heat energy released when 29.5 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point is the sum of the heat energy for the temperature change and the heat energy for the phase change: 868.79 J + 341.18 J = 1209.97 J.