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A 348 g metal piece absorbs 6.64 kJ and warms from 22.4 degrees C to 43.6 degrees C. What is its specific heat?

A0.900 J/g degree C
B0.449 J/g degree C
C1.11 J/g degree C
D0.129 J/g degree C
Answer & Solution
Correct answer: A. 0.900 J/g degree C
1. Rearrange q equals c times m times the temperature change to give c equals q over m times the change. 2. Convert the heat: 6.64 kJ is 6640 J. 3. The temperature change is 43.6 minus 22.4, which is 21.2 degrees Celsius. 4. The denominator is 348 g times 21.2 degrees Celsius, which is 7377.6 gram degrees Celsius. 5. Divide: 6640 divided by 7377.6 gives 0.900 J per gram per degree Celsius. 6. Comparing with tabulated values, this matches aluminium at 0.897 J/g degree C. 7. Leaving the heat in kilojoules would give 0.000900 and misplace the decimal point entirely. _Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.1 Energy Basics_
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