Question

Calculate the amount of energy required to boil 25.00 g of mercury.

Answer 1

Answer:

C. 7.4 kJ

Explanation:

Let assume that Mercury is at room temperature (25 °C). The energy required to boil the sample of mercury is the sum of sensible and latent heats. Mercury has a fussion and boiling points of -38.83 °C and 356.7 °C, respectively, and a specific heat of $138\,\frac{J}{kg\cdot ^{\circ}C}$. Then:

$Q = (25\,g)\cdot \left[\left(0.138\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (356.7^{\circ}C - 25^{\circ}C) + 296\,\frac{J}{g} \right]$

$Q = 8544.365\,J$

The option that offers the best approximation to the result is C.