# Calculate the amount of heat that must be released to convert 20.0 g of mercury vapor at 387^@"C" to liquid mercury at 307^@"C" (in kJ)?

Jun 6, 2015

Heat added or removed from a substance when there is a change of temperature without a change in phase ($H$) $= m . s . \Delta t$

where $m$ is mass of substance
$s$ is specific heat
$\Delta t$ is absolute change in temperature

However, we need to appreciate the underlying steps in the thermal transformation taking place. The change takes place in 3 steps:

1. Extraction of heat to lower temperature of vapour from ${387}^{o} C$ to ${357}^{o} C$ (phase transition temperature).
2. Phase transition from vapour to liquid without change in temperature ( Heat added/removed = Heat of phase transition/unit mass * mass of substance )
3. Extraction of heat to lower temperature of liquid from ${357}^{o} C$ to ${307}^{o} C$.

Step I: ${H}_{1} = 20 \cdot 0.104 \cdot \left(387 - 357\right)$ Vapour
$\implies {H}_{1} = 20 \cdot 0.104 \cdot 30 = 62.4 J$

Step II: ${H}_{2} = 292 \cdot 20 = 5840 J$ Phase transition

Step III: ${H}_{3} = 20 \cdot 0.138 \cdot \left(357 - 307\right)$ Liquid
$\implies {H}_{3} = 20 \cdot 0.138 \cdot 50 = 138 J$

Total heat removed = ${H}_{1} + {H}_{2} + {H}_{3} = 62.4 + 5840 + 138 J = 6.040 k J$