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

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point = 34.60°C Hvap(34.60°C) = 357.5 J/g melting point = -116.3°C
Hfus(-116.3°C) = 98.10 J/g specific heat gas = 1.460 J/g⋅°C
specific heat liquid = 2.320 J/g⋅°C A 27.70 g sample of liquid
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The following information is given for bismuth
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boiling point =
1.627E3 °C
Hvap(1.627E3°C) =
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melting point =
271.0 °C
Hfus(271.0°C) = 52.60
J/g
specific heat solid=
0.1260 J/g ⋅ °C
specific heat liquid =
0.1510 J/g ⋅ °C
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into a mold and allowed to cool to 24.0 °C. How
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Tb =
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Hvap(36.20°C) = J/g
Tm=
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Hfus(–129.70°C) = J/g
Specific heat gas =
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Specific heat liquid =
2.280 J/g °C
A 47.80 g sample of liquid
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Delta H vaporization(78.40°C) = 837.0 J/g
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Delta H fusion(–114.50°C) = 109.0 J/g
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specific heat solid = 0.2300 J/g°C
specific heat liquid = 0.2640 J/g°C
A 20.60 g sample of liquid cadmium at 375.0 °C is poured into a
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n-pentane at 1 atm:
= 36.20°C
(36.20°C)
= J/g
=
–129.70°C
(–129.70°C)
= J/g
Specific heat gas =
1.650 J/g °C
Specific heat liquid =
2.280 J/g °C
A 40.00 g sample of liquid
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