- #1

krootox217

- 51

- 2

## Homework Statement

The heat capacity of the solid NaCl from 500 K to 1074 K is given by [52.996 J*K

^{-1}*mol

^{-1}– (7.86*10

^{-3}J*K

^{-2}*mol

^{-1})*T + (1.97*10

^{-5}J*K

^{-3}*mol

^{-1})*T2 ] and that of liquid NaCl from 1074 K to 1500 K is given by [125.637 J*K

^{-1}mol

^{-1}– (8.187*10

^{-2}J*K

^{-2}*mol

^{-1})*T + (2.85*10

^{-5}J*K

^{-3}*mol

^{-1})*T2 ]. If Δfus H

^{Θ}

_{T, m}= 28.158 kJ*mol

^{-1}at 1074K, Please determine H

^{Θ}1500K - H

^{Θ}500K for NaCl.

## Homework Equations

## The Attempt at a Solution

**I tried to solve this task, but I don't understand what they exactly want me to do.**

**Do I have to calculate the ΔH from 500K to 1074K with the given heat capacities? And what is Δ**

_{fus}H^{Θ}?