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A 59.7 g metal from boiling water at 100.0 degrees C is put into 60.0 g of water at 22.0 degrees C, reaching 28.5 degrees C. What is the metal?

ACopper, specific heat about 0.38 J/g degree C
BAluminium, specific heat about 0.90 J/g degree C
CLead, specific heat about 0.13 J/g degree C
DHelium, specific heat about 5.19 J/g degree C
Answer & Solution
Correct answer: A. Copper, specific heat about 0.38 J/g degree C
1. Heat gained by the water is 4.184 J/g degree C times 60.0 g times 6.5 degrees Celsius. 2. That product is 1631.8 times 6.5, which equals 1606.8 J. 3. The metal cooled from 100.0 to 28.5 degrees C, a change of minus 71.5 degrees Celsius. 4. Its denominator is 59.7 g times minus 71.5, which is minus 4268.6 gram degrees Celsius. 5. The metal heat is minus 1606.8 J, so dividing gives 0.376 J per gram per degree Celsius. 6. Rounded, that is about 0.38 J/g degree C. 7. The tabulated value nearest to it is copper, so the metal is identified as copper. 8. Aluminium and lead are far away at 0.897 and 0.130 J/g degree C. _Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.2 Calorimetry_
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