The following information is given for lead at 1atm: specific heat liquid = 0.13

Question

The following information is given for lead at 1atm:

specific heat liquid = 0.1380 J/g°C

A 48.50 g sample of solid lead is initially at 317.0 °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 626.0 °C.

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 48.50 g sample of solid lead is initially at 317.0 °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 626.0 °C.

Explanation / Answer

Solid Sensible heat

Q1 = m*Cp*(Tf-Ti) = 48.5*0.130*(328-317) = 69.355 J

Latent heat of melting

Q2 = m*LH = 48.5*23 = 1115.5 J

Snesible heat of liquid

Q3 = m*Cpl*(Tf-Ti) = 48.5*0.138(626-328) =1994.514J

QT = Q1+Q2+Q3 = 69.355 + 1115.5 +1994.514J= 3179.369 J

Qt = 3.17 kJ

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.