Question

A 20.0-mL sample of benzene at 21.4 °C was cooled to its melting point, 5.5 °C, and then frozen. How much energy was given off as heat in this process? (The density of benzene is 0.80 g/mL, its specific heat capacity is 1.74 J/g · K, and its heat of fusion is 127 J/g.)

Answer 1

#76279 Chemistry, General Chemistry

A 20.0 mL sample of benzene at 21.4°C was cooled to its melting point, 5.5°C, and then frozen. How much energy was given off as heat in this process? (The density of benzene is 0.80 g/mL, its specific heat capacity is 1.74 J/g · K, and its heat of fusion is 127 J/g).

Answer:

$$\begin{array}{l} \rho = m / V \quad m = \rho \cdot V \\ m \text{ (benzene sample)} = 25.0 \, \text{mL} \cdot 0.80 \, \text{g/mL} = 20.0 \, \text{g} \\ \text{Specific heat capacity for benzene indicates that cooling 1 g of benzene by 1 K requires 1.74 J.} \\ \text{Therefore cooling 20.0 g of benzene by 14.4 K requires the removal of energy: } 20.0 \cdot 14.4 \cdot 1.74 \, \text{J} = 501.12 \, \text{J} \\ \text{As this is loss of energy by the system, it has a negative sign (-501.12 J).} \end{array}$$

Heat of fusion for benzene indicates that 1 g of benzene loses 127 J to freeze. That is why, 20.0 g loses:

$$20 \times 127 \, \text{J} = 2540 \, \text{J}$$

As this is also loss of energy by the system, it also has a negative sign (-2540 J).

Total energy lost: -501.12 + (-2540 J) = 3041 J

Processes that are exothermic (give out heat) have a negative sign.

Processes that are endothermic (absorb heat) have a positive sign.