A 698 g sample of a substance increases in temperature by 2550C when 32,000 J of energy is added. What is the specific heat of the substance?Group of answer choices125 J/g ºC45.8 J/g ºC0.180 J/g ºC0.296 J/g ºC
Question
A 698 g sample of a substance increases in temperature by 2550C when 32,000 J of energy is added. What is the specific heat of the substance?Group of answer choices125 J/g ºC45.8 J/g ºC0.180 J/g ºC0.296 J/g ºC
Solution
The specific heat of a substance can be calculated using the formula:
q = mcΔT
where: q = heat energy absorbed or released (in joules, J) m = mass of the substance (in grams, g) c = specific heat capacity (in joules per gram per degree Celsius, J/g°C) ΔT = change in temperature (in degrees Celsius, °C)
We can rearrange the formula to solve for c:
c = q / (mΔT)
Substituting the given values:
c = 32,000 J / (698 g * 2550°C)
c = 32,000 J / 1,779,900 g°C
c = 0.018 J/g°C
However, none of the provided answer choices match this result. There may be a mistake in the question or the answer choices.
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