# A 29.80 g sample of liquid cadmium at 377.00°C is poured into a mold and allowed to cool to 24.00°C. How many kJ of energy are released in this process?

## The following information is given for cadmium at 1atm: Tb = 765.00°C Hvap (765.00°C) =889.6 J/g Tm = 321.00°C Hfus (321.00°C) = 54.40 J/g Specific heat solid = 0.2300 J/g °C Specific heat liquid = 0.2640 J/g °C (Report the answer as a positive number.) Energy= kJ

Nov 8, 2017

The total energy released is 4097 J (to four sig. digits) or 4.097 kJ

#### Explanation:

Since you calcium starts this process at 377 °C and is in liquid phase, the heat of vapourization does not enter into this. However, the heat of fusion does, as the calcium will freeze into a solid before it reaches 24 °C

So, three things happen, each of which involves a quantity of heat.

First, the liquid calcium cools from 377 °C to 321 °C. Since the specific heat for the liquid is 0.2640 J/g °C, this heat is

$E = m \times C \times \Delta t$

where $m$ is the mass, $C$ is the specific heat and $\Delta t$ is the temperature change.

E = 29.80g xx 0.2640 J/(g °C) xx 56 °C = 440.56 J

Second, the liquid freezes at 321 °C:.

$E = 29.80 g \times 54.40 \frac{J}{g} = 1621.12 J$

Finally, the solid cools. The calculation is similar to the first one above:

E = 29.80g xx 0.2300 J/(g °C) xx 297 °C = 2035.64J

(notice the change to the specific heat of the solid, and the large drop in temperature, from 321 ° to 24 °)

Total energy is the sum: $440.56 + 1621.12 + 2035.64 = 4097.32 J$