The following information is given for lead at 1 atm: specific heat liquid = 0.1

Question

The following information is given for lead at 1 atm:

specific heat liquid = 0.1380 J/g°C

A 30.00 g sample of liquid lead at 580.0 °C is poured into a mold and allowed to cool to 28.0 °C. How many kJ of energy are released in this process. Report the answer as a positive number.

kJ

boiling point = 1.740×103 °C Hvap(1.740×103 °C) = 858.2 J/g melting point = 328.0 °C Hfus(328.0 °C) = 23.00 J/g specific heat solid = 0.1300 J/g°C

specific heat liquid = 0.1380 J/g°C

A 30.00 g sample of liquid lead at 580.0 °C is poured into a mold and allowed to cool to 28.0 °C. How many kJ of energy are released in this process. Report the answer as a positive number.

kJ

Explanation / Answer

Q1 = m*Cliq*(Tf-Ti)

Q2 = m*Hfus

Q3 = m*Cpsolid*(T2-Tf)

then

Q1 = 30*(0.1380 )*(328-580) = -1043.28

Q2 = 30*-23 = -690

Q3 = 30*0.13*(28-328) = -1170

QT = -(1043.28+690+1170) = -2903.28 J

QT =  -2903.28 J = -2.9 kJ

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