Question

Calculate the heat flow when 13.8 G of benzene at 10.0 degrees C is cooled to -10.0 Celsius

Answer 1

Answer on Question #84727 – Chemistry – Other

Task:

Calculate the heat flow when 13.8 g of benzene at 10.0°C is cooled to -10.0°C.

Solution:

Benzene

Thermodynamic properties

Benzene freezes at 5.5°C = 278.65 K (melting point of benzene).

Benzene liquid at 10°C.

Benzene solid at -10°C.

Heat capacity for liquid benzene = 134.8 J/mol*K.

Heat capacity for solid benzene = 118.4 J/mol*K.

Enthalpy change of fusion for benzene = 9.9 kJ/mol.

Molar mass of benzene = 78.11 g/mol.

Convert these values from J/mol*K to J/K:

Heat capacity for liquid benzene = C(C₆H₆, liquid) = (134.8 J mol⁻¹ K⁻¹) / (78.11 g mol⁻¹) = 1.726 J/K = 1.726 J/°C;

Heat capacity for liquid benzene = 1.726 J/°C.

Heat capacity for solid benzene = C(C₆H₆, solid) = (118.4 J mol⁻¹ K⁻¹) / (78.11 g mol⁻¹) = 1.516 J/K = 1.516 J/°C;

Heat capacity for solid benzene = 1.516 J/°C.

Enthalpy change of fusion for benzene = ΔfusH° = (9900 J mol⁻¹) / (78.11 g mol⁻¹) = 126.744 J/g;

Enthalpy change of fusion for benzene = 126.744 J/g.

Liquid benzene will release some of its internal energy to the surroundings when it cools and then solidifies. Solid benzene will also release some of its internal energy when it cools.

1) lowering liquid benzene temperature from 10°C to 5.5°C;

2) freezing the benzene;

3) cooling the solid benzene from 5.5°C to -10°C.

1) Cooling liquid benzene:

$$\Delta Q = C m \Delta T$$

Where:

ΔQ – amount of heat; C - specific heat capacity; m – material mass; ΔT – temperature rise.

Then,

$$\Delta Q _ {1} = C m \Delta T = (1.726 \mathrm {J} / \mathrm {g} \cdot {} ^ {\circ} \mathrm {C}) * (13.8 \mathrm {g}) * (5.5 {} ^ {\circ} \mathrm {C} - 10 {} ^ {\circ} \mathrm {C}) = - 107.1846 \mathrm {J}$$ $$\Delta Q _ {1} = - 107.1846 \mathrm {J}$$

2) Solidification:

$$\Delta Q = m * L$$

Where:

ΔQ – amount of heat; L = -ΔfusH° - heat of fusion; m – material mass.

Then,

$$\Delta Q _ {2} = m * L = (13.8 \mathrm {g}) * (- 126.744 \mathrm {J} / \mathrm {g}) = - 1749.0672 \mathrm {J}$$ $$\Delta Q _ {2} = - 1749.0672 \mathrm {J}$$

3) Cooling solid benzene:

$$\Delta Q = C m \Delta T$$

Where:

ΔQ – amount of heat; C - specific heat capacity; m – material mass; ΔT – temperature rise.

Then,

$$\begin{array}{l} \Delta Q _ {3} = C m \Delta T = (1.516 \mathrm{J}/\mathrm{g} \cdot^{\circ} \mathrm{C}) * (13.8 \mathrm{g}) * (-10^{\circ} \mathrm{C} - 5.5^{\circ} \mathrm{C}) = -324.2724 \mathrm{J} \\ \Delta Q _ {3} = -324.2724 \mathrm{J} \end{array}$$

Finally, the total energy released is:

$$\begin{array}{l} \Delta Q = \Delta Q _ {1} + \Delta Q _ {2} + \Delta Q _ {3} = (-107.1846 \mathrm{J}) + (-1749.0672 \mathrm{J}) + (-324.2724 \mathrm{J}) = -2180.5242 \mathrm{J} \\ \Delta Q = -2180.5 \mathrm{J} \approx -2.18 \mathrm{kJ} \end{array}$$

The negative sign represents energy lost by the system.

**Answer:** Heat flow = $-2.18 \mathrm{kJ}$.

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