Question
A 28.2 g sample of beryllium at 96.7C is placedinto 67.5 mL of water at 20.2C in an insulatedcontainer. The temperature of the water at
A 28.2 g sample of beryllium at 96.7°C is placedinto 67.5 mL of water at 20.2°C in an insulatedcontainer. The temperature of the water at thermal equilibrium is32.0°C. What is the specific heat of beryllium? Assume a density of1.00 g/mL for water. (See this table.)
J/(g·°C)
Constants Used In This Book (experimentalvalues)
| 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 |
Unit Conversion Factors (bydefinition)
| 0 | °C | = | 273.15 | K |
| 1 | foot | = | 12 | inches |
| 1 | mile | = | 5280 | feet |
| 1 | inch | = | 2.54 | cm |
| 1 | mile | = | 1.609 | km |
| 1 | Å | = | 1×10-10 | m |
| 1 | cal | = | 4.184 | J |
| 1 | Cal | = | 1000 | cal |
| 1 | eV | = | 1.602×10-19 | J |
| 1 | eV | = | 96.485 | kJ/mol |
| 1 | L·atm | = | 101.3 | J |
| 1 | lb | = | 453.6 | g |
| 1 | kg | = | 2.205 | lbs |
| 1 | N | = | 1 | kg·m/s2 |
| 1 | J | = | 1 | N·m |
| 1 | Pa | = | 1 | N/m2 |
| 1 | atm | = | 14.7 | psi |
| 1 | atm | = | 760 | mmHg |
| 1 | atm | = | 760 | torr |
| 1 | atm | = | 1.01325 | bar |
| 1 | atm | = | 1.01325×105 | Pa |
| 1 | D | = | 3.34×10-30 | C·m |
| 1 | Ci | = | 3.7×1010 | s-1 |
| 1 | Gy | = | 1 | J/kg |
| 1 | Gy | = | 100 | rad |
| 1 | R | = | 2.58×10-4 | C/kg |
| 1 | yr | = | 365 | d* |
Mathematical Constants (infiniteprecision)
| ? | = | 3.14159265359 |
| e | = | 2.71828182846 |
| ln(2) | = | 0.69314718056 |
| ln(10) | = | 2.30258509299 |
Standard Conditions
| STP | (standard temperature and pressure) | = | 0°C and 1 atm |
| SATP | (standard ambient temperature and pressure) | = | 25°C and 1 bar |
Step by Step Solution
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At thermal equilibrium in an insulated container heat lost by ...
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Chemical Principles
Authors: Steven S. Zumdahl, Donald J. DeCoste
7th edition
9781133109235, 1111580650, 978-1111580650
