Question

Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. The average specific heat of gaseous ethanol is about 1.80 J/g-K. a. How much heat is required to convert 35.0 g of ethanol at 27 °C to the vapor phase at 120 °C? b. How much heat is required to convert the same amount of ethanol at –120 °C to the vapor phase at 120 °C?

Answer 1

To calculate the heat required for ethanol phase changes at different temperatures, separate calculations for heating, melting, boiling, and further heating must be performed and then summed together for the total heat.

Step-by-step explanation:

To calculate the heat required to convert ethanol from one phase to another, it's important to consider the phase changes and temperature changes separately. We'll calculate the heat for each step and then add them together to find the total heat required.

Converting Ethanol at 27 ℃ to Vapor at 120 ℃

First, we calculate the heat required to raise the temperature of liquid ethanol from 27 ℃ to its boiling point at 78 ℃. We use the formula q = m × c × ΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.

Next, we calculate the heat of vaporization to convert liquid ethanol at its boiling point to vapor using the formula q = n × ΔHvap, where n is the number of moles and ΔHvap is the enthalpy of vaporization.

Lastly, we calculate the heat required to raise the temperature of the gaseous ethanol from 78 ℃ to 120 ℃. Again, we use the formula q = m × c × ΔT.

By adding the heat values calculated from these steps, we get the total heat required for the ethanol to reach the vapor phase at 120 ℃.

Converting Ethanol at -120 ℃ to Vapor at 120 ℃

The calculation for converting ethanol at -120 ℃ includes additional steps for heating the ethanol from its solid state at -120 ℃ to its melting point at -114 ℃, and the heat of fusion to melt the solid into a liquid.

Answer 2

First question

`Q = 36826 \ J`

Second question

`Q = 52299.7 \ J`

Step-by-step explanation:

From the question we are told that

The melting point of Ethanol is
`T_m = -114 ^oC`

The boiling point of Ethanol is
`T_b = 78^ oC`

The enthalpy of fusion of Ethanol is
`F = 5.02 \ kJ / mol = 5.02 *10^(3)\ kJ / mol`

The enthalpy of vaporization of Ethanol is
`L = 38.56 \ kJ / mol = 38.56 *10^(3) \ J / mol`

The specific heat of solid Ethanol is
`c_e = 0.97 \ J/ g \cdot K`

The specific heat of liquid Ethanol is
`c_l = 2.3 \ J / g \cdot K`

The mass of the Ethanol given is
`m = 35.0 \ g`

Considering the first question

The initial temperature is
`T_i = 27^oC`

The final temperature is
`T_f = 120^oC`

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

`Q_1 = m * c_l * (T_b - T_i)`

=>
`Q_1 = 35.0 * 2.3 * ( 78 - 27)`

=>
`Q_1 =4106 \ J`

Genially the number of moles of Ethanol given is mathematically represented as

`n = (m)/(Z)`

Here Z is the molar mass of Ethanol with value
`Z = 46 g/mol`

So

`n = (35)/(46 )`

=>
`n = 0.7609 \ mol`

Generally the heat of vaporization of the Ethanol is mathematically represented as

`Q_2 = n * L`

=>
`Q_2 =0.7809 * 38.56 * 10^(3)`

=>
`Q_2 =29339 \ J`

Generally the heat required too raise the Ethanol from its boiling point to
`T_f` is mathematically represented as

`Q_3 = m * c_l * (T_f - T_b)`

=>
`Q_3 = 35 * 2.3 * (120 - 78 )`

=>
`Q_3 = 3381 \ J`

Generally the total heat required is

`Q = Q_1 + Q_2 + Q_3`

=>
`Q = 4106 + 29339 + 3381`

=>
`Q = 36826 \ J`

Considering the second question

The initial temperature is
`T_i = -120^oC`

The final temperature is
`T_f = 120^oC`

Generally the heat required too raise the Ethanol to its melting point is mathematically represented as

`Q_1 = m * c_e * (T_m - T_i)`

=>
`Q_1 = 35.0 * 0.97 * ( -114 - (- 120) )`

=>
`Q_1 = 203.7 \ J`

Generally the heat of fusion of the Ethanol is mathematically represented as

`Q_2 = n * F`

=>
`Q_2 =0.7809 * 5.02 *10^(3)`

=>
`Q_2 =3920 \ J`

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

`Q_3 = m * c_l * (T_b - T_m)`

=>
`Q_3 = 35.0 * 2.3 * ( 78 - (- 114) )`

=>
`Q_3 =15456 \ J`

Generally the heat of vaporization of the Ethanol is mathematically represented as

`Q_4 = n * L`

=>
`Q_4 =0.7809 * 38.56 * 10^(3)`

=>
`Q_4 =29339 \ J`

Generally the heat required too raise the Ethanol from its boiling point to
`T_f` is mathematically represented as

`Q_5 = m * c_l * (T_f - T_b)`

=>
`Q_5 = 35 * 2.3 * (120 - 78 )`

=>
`Q_5 = 3381 \ J`

Generally the total heat required is

`Q = Q_1 + Q_2 + Q_3+Q_4 + Q_5`

=>
`Q = 203.7 + 3920 + 15456 +29339+3381`

=>
`Q = 52299.7 \ J`