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The temperature of a 12.58 g sample of calcium carbonate [CaCO3 (s)] increases from 23.6 °C to 38.2 °C. If the specific heat of calcium carbonate is 0.82 J/g-K, how many joules of heat are absorbed?

Answer :

Final answer:

To calculate the amount of heat absorbed, use the formula Q = m * c * ΔT, where Q is the heat absorbed, m is the mass of the sample, c is the specific heat capacity of the substance, and ΔT is the change in temperature. Given the values in the question, we find that approximately 150.319 joules of heat are absorbed.

Explanation:

To calculate the amount of heat absorbed, we can use the formula:

Q = m * c * ΔT

Where:
m = mass of the sample (in grams)
c = specific heat capacity of the substance (in J/g-°C)
ΔT = change in temperature (in °C)

Given:

  • Mass of sample = 12.58 g
  • Specific heat of calcium carbonate = 0.82 J/g-°C
  • Change in temperature = 38.2 °C - 23.6 °C = 14.6 °C

Plugging the values into the formula, we get:

Q = 12.58 g * 0.82 J/g-°C * 14.6 °C

Simplifying, we find:

Q ≈ 150.319 J

Therefore, approximately 150.319 joules of heat are absorbed.

Learn more about calcium carbonate here:

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