A 28.2 g sample of beryllium at 96.7°C is placed into 67.5 mL of water at 20.2°C

Question

A 28.2 g sample of beryllium at 96.7°C is placed into 67.5 mL of water at 20.2°C in an insulated container. The temperature of the water at thermal equilibrium is 32.0°C. What is the specific heat of beryllium? Assume a density of 1.00 g/mL for water. (See this table.)
J/(g·°C)

Constants Used In This Book (experimental values)

Unit Conversion Factors (by definition)

Mathematical Constants (infinite precision)

Standard Conditions

Constant Value Units Usage nAv 6.0221×1023 particles Avogadro's number R 0.08206 L·atm/K·mol ideal gas constant R 8.3145 J/K·mol ideal gas constant R 1.9872 cal/K·mol ideal gas constant Vm 22.414 L/mol ideal gas law @ 0°C Vm 24.465 L/mol ideal gas law @ 25°C c 2.9979×108 m/s speed of light g 9.81 m/s2 standard gravity h 6.6261×10-34 m2/kg·s Planck's constant a0 5.2918×10-11 m Bohr radius k 1.3807×10-23 J/K Boltzmann's constant C 6.2415×1018 e Coulomb's Law F 96485 C/mol Faraday's constant R 1.097×107 m-1 Rydberg constant RH 2.18×10-18 J Rydberg constant of hydrogen amu 1.6605×10-27 kg atomic mass unit mass e-, e+ 9.1094×10-31 kg elementary particle mass p+ 1.6726×10-27 kg elementary particle mass n0 1.6749×10-27 kg elementary particle charge e 1.6022×10-19 C elementary particle energy En 931.5 MeV elementary particle Nernst factor 0.0592 --- Nernst equation Kw 1.0×10-14 --- autoionization constant of water Ka,H2O 1.8×10-16 --- acidity constant of water

Explanation / Answer

At thermal equilibrium in an insulated container

, heat lost by beryllium = heat gained by water

Mass of beryllium*specific heat of beryllium * temperature difference= mass of water* specific heat of water * temperature difference

=28.2*specific heat of beryllium * (96.7-32)= 67.5*1*4.18*(32-20.2)

Specific heat of beryllium =1.824 j/g.deg.c

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