chemistry heat transfer help!! SearchShare More>» Use the References to access i

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chemistry heat transfer help!! SearchShare More>» Use the References to access important values if needed for this question. The following information is given for lead at latm: boiling point - 1.740 103 °c melting point 328.0°C specific heat solid = 0.1300 J/g°C specific heat liquid = 0.1380 JgoC ,p(1.740-103 oC)-858.2 J/g ARins(328.0 °C) = 23.00 J/g boiling poin-1.740x 103 oC A 24.10 g sample of liquid lead at 578.0 °C is poured into a mold and allowed to cool to 29.0 °C. How many kJ of energy are released in this process. Report the answer as a positive number kJ Submit Answer Retry Entire Group 6 more group attempts remaining

Explanation / Answer

The lead sample is below the boiling point. Cooling it from 578o C to 29o C involves three steps.

1. Cooling from 578o C(liquid) to 328o C (liquid).

Heat released = m.Cm.dT = 24.10 g * 0.1380 J/goC * (578-328)oC

= 831.45 J

2. Cooling from 328o C (liquid) to 328o C (solid).

Heat released = m.Hfus = 24.10g * 23 J/g

= 554.3 J

3. Cooling from 328o C (solid) to 29o C (solid).

Heat released = m.Cm.dT = 24.10 g * 0.1300 J/goC * (328-29)oC

= 936.767 J

Total heat released

= 831.45 J + 554.3 J + 936.767 J

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