Question

The following information is given for cobalt at 1 atm: boiling point = 3.097E3°C Hvap(3.097E3°C) = 6.603E3 J/g melting point = 1.495E3°C Hfus(1.495E3°C) = 262.7 J/g specific heat solid = 0.4180 J/g ⋅ °C specific heat liquid = 0.6860 J/g ⋅ °C A 38.00 g sample of solid cobalt is initially at 1.467E3°C. If the sample is heated at constant pressure (P = 1 atm), __ kJ of heat are needed to raise the temperature of the sample to 1.963E3°C.

Answer 1

Ti = 1467.0 oC

Tf = 1963.0 oC

here

Cs = 0.418 J/g.oC

Heat required to convert solid from 1467.0 oC to 1495.0 oC

Q1 = m*Cs*(Tf-Ti)

= 38 g * 0.418 J/g.oC *(1495-1467) oC

= 444.752 J

Lf = 6603.0 J/g

Heat required to convert solid to liquid at 1495.0 oC

Q2 = m*Lf

= 38.0g *6603.0 J/g

= 250914 J

Cl = 0.686 J/g.oC

Heat required to convert liquid from 1495.0 oC to 1963.0 oC

Q3 = m*Cl*(Tf-Ti)

= 38 g * 0.686 J/g.oC *(1963-1495) oC

= 12199.824 J

Total heat required = Q1 + Q2 + Q3

= 444.752 J + 250914 J + 12199.824 J

= 263559 J

= 264 KJ

Answer: 264 KJ