Question

Please explain with the steps....

Thanks

For Fe (Iron) at 1 atm:

Boiling point: 2750.0 C

Heat of vaporization: 1.51 x 10^3 cal/g

Heat of fusion: 69.1 cal/g

Melting point: 1535.0 C

In order to freeze it at its normal melting point of 1535.0°C, How many kcal of energy must be removed from a 35.6 g sample of liquid iron????

Energy removed is .................... in kcal?

Answer 1

Q = Heat of fusion $\times$ mass

$Q = 69.1 \times 35.6 = 2460 \;cal$

Energy removed is 2.46 in kcal