Constants I Periodic TableThe specific heat of mercury is 138 J/kg · C°.Part ADetermine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting pointof -39.0 °C is placed in a 0.620-kg aluminum calorimeter with 0.400 kgtemperature is 5.06 °C.water at 12.80 °C; the resulting equilibriumExpress your answer with the appropriate units.Lfusion =ValueUnitsSubmitRequest AnswerProvide FeedbackNext >P PearsonCopyright © 2020 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | Contact Us |MacBook Pro

Given Mass of mercury mm = 1.0 kgSpecfic heat of mercury Cm =138 J/kg.°CMelting point (T) = -39.0°CEquilibrium temperature(T1) = 5.06°CMass of water mw = 0.400 kgSpecfic heat of water Cw = 4186 J/kg.°CTemperature of water T2 = 12.80°CMass of aluminum ma = 0.620 kgSpecfic heat of water Ca = 900J/kg.°CHere, the latent heat of fusion of mercury is determined from conservation of energy is as follows: mmL + mmCm T-T1=mwCw T2-T+maCa T2-TmmCm T-T1 = 1.0 kg 138 J/kg.°C5.06°C--39.0°C = 6080.28 JmwCw T2-T =0.400 kg 4186 J/kg.°C 12.80°C-5.06°C = 12960 JmaCa T2-T =0.620 kg 900 J/kg.°C 12.80°C-5.06°C = 4319 J Hence, the latent heat of fusion of mercury becomes as, Lf = mwCw T2-T+maCa T2-T- mmL + mmCm T-T1mm = 12960 J+4319J-6080.28J1.0 kg =11199 J/kgLf = 1.12×104 J/kg

Answered: Constants I Periodic Table The specific…