Question

A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature rises from 25.0 to 62.3°C, determine the heat capacity of the calorimeter. The molar mass of ethanol is 46.07 g/mol. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔH°rxn = -1235 kJ

Answer 1

The heat capacity of the calorimeter in which 21.8 g of ethanol is burned is 15661 J/ºC

We'll begin by calculating the number of mole of ethanol. This can be obtained as follow:

Mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Mole of ethanol =?

Mole = mass / molar

Mole of ethanol = 21.8 / 46.07

Mole of ethanol = 0.473 mole

Next, we shall determine the heat released. This can be obtained as follow as:

Mole of ethanol = 0.473 mole

Enthalpy change (ΔH°rxn) =  –1235 kJ/mol = –1235 × 1000 = –1235000 J/mol

Heat released (Q) =?

Q = n × ΔH

Q = 0.473 × –1235000

Q = –584155 J

Thus, the heat released is –584155 J

NOTE:

Heat release by ethanol = heat absorbed by calorimeter

Finally, we shall determine the heat capacity of the calorimeter.

Initial temperature (T₁) = 25 °C

Final temperature (T₂) = 62.3 °C

Change in temperature (ΔT) = 62.3 – 25 = 37.3 °C

Heat absorbed (Q) =  584155 J

Heat capacity (C) =?

Q = CΔT

584155 = C × 37.3

Divide both side by 37.3

C = 584155 / 37.3

C = 15661 J/ºC

Therefore, the heat capacity of the calorimeter is 15661 J/ºC

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Answer 2

The specific heat capacity of combustion of 21.8 of ethane is $\rmm \bold { 718. 394 J/kg^ \cdot C}$

Given here,

The molar mass of ethanol is 46.07 g/mol.

Enthalpy of the combustion $\rm \bold{ \Delta H = -1235 kJ}$

Hence the number of moles in 21.8 g of ethanol is 0.473 mole

The energy  released during reaction

$\rm \bold{ Q = n \times \Delta H}$

Q = - 584.15 kJ

As we know Heat released by ethanol is equal to hat absorbed by calorimeter.

The heat absorbed = 584.15 kJ

The specific heat capacity formula

$\rm \bold{ c= \frac{Q}{m \times \Delta T} }$

Where,

Q = heat absorbed =  584155 J

m = mass = 21.8g

$\rm \bold{ \Delta T} }$- change in temperature =  37.3 °C

$\rm \bold{ c= \frac{584.15 kJ}{ 21.8g \times 37.3} }$

$\rmm \bold { c = 718. 394 J/kg^ \cdot C}$

Hence, we can conclude that the specific heat capacity of combustion of 21.8 of ethane is $\rmm \bold { 718. 394 J/kg^ \cdot C}$

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