Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,
[tex]Q = m s dT[/tex]
Where, m is mass of iron
s is specific heat of iron
d T is change in temperature in celcius
[tex]= 15.6 g * 0.45 J /g /C * (355 - 122) = 1.63 * 10^3 J[/tex]
If
1 cal = 4.2 J
Then,
[tex]Q = (1.63 * 10^3) /4.2 = 0.389 K cal[/tex]
Thus 0.389 k cal of enrgy is required by a 15.6 g Fe to reach to 355 C^0
