Question

Use the References to access impoPlaNE VAIDOS II needed for this question. The following information is given for cobalt at 1 atm: AHvap(3097.00°C) = 6.603 x 10' J/g AH fus( 1495.00°C) = 262.7 J/g Tb = 3097.00°C Tm = 1495.00°C Specific heat solid = 0.4180 J/g °C Specific heat liquid = 0.6860 J/g °C A 30.50 g sample of solid cobalt is initially at 1472.00°C. If 1.053 x 10* J of heat are added to the sample at constant pressure (P= 1 atm), which of the following is/are true? (Select all that apply.) The sample is a liquid. The sample is at a temperature of 1495.00°C. The sample is at a temperature greater than 1495.00°C. The sample is a solid in equilibrium with liquid. The sample is a gas. Submit Answer

Answer 1

1) Calculate the heat required to increae temperature of 30.50 g of solid cobalt from 1472.00 deg C to 1495.00 deg C (melting point):

q= mass of solid cobalt x specific heat of solid x temperature change q = 30.50 g x 0.4180 x ( 1495.00 °C – 1472.00 °C) q= 293.23 J

2) Calculate the heat required to melt 30.50 g of solid cobalt at 1495.00 deg C (melting point):

q= mass of solid cobalt X AHfus 9 = 30.50 g x 262.7 - q= 8012.35 J

3) Calculate the heat required to increae temperature of 30.50 g of liquid cobalt from 1495.00 deg C (melting point) to 3097.00 deg C (boiling point)::

q= mass of liquid cobalt x specific heat of liquid x temperature change 9 = 30.50 g x 0.6860 — X ( 3097.00 °C 1495.00 °C) 9.CX (309 q= 33469.79 J

Add the heat required to increae temperature of 30.50 g of solid cobalt from 1472.00 deg C to 1495.00 deg C (melting point) to the heat required to melt 30.50 g of solid cobalt at 1495.00 deg C (melting point):

293.23 J + 8012.35 J = 2224.42 J

Add the heat required to increae temperature of 30.50 g of solid cobalt from 1472.00 deg C to 1495.00 deg C (melting point) to the heat required to melt 30.50 g of solid cobalt at 1495.00 deg C (melting point) and to the heat required to increae temperature of 30.50 g of liquid cobalt from 1495.00 deg C (melting point) to 3097.00 deg C (boiling point):

293.23 J + 8012.35 J + 33469.79 J = 33469.79 J

Total heat given to obat sample is
1.053 x 104]
. This is grater than
2224.42 J
but less than
33469.79 J
.

Hence, the sample is liquid and the final teperature is greater than 1495.00 deg C (melting point) but less than 3097.00 deg C (boiling point)::