A 500.0 g sample of aluminum, initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. 32.62245 kJ heat has been absorbed by the sample.
What is specific heat?The term specific heat is defined as the amount of heat required to increase the temperature of 1 gram of a substance 1 degree Celsius (°C).
To calculate the amount of heat absorbed by the sample, use the formula:
Q = mcΔT
where Q is the amount of heat absorbed by the sample, m is the mass of the sample, c is the specific heat of aluminum, and ΔT is the change in temperature of the sample.
Substituting the given values into the formula, we get:
Q = 500.0 g × 0.9930 J/g·K × (90.7°C - 25.0°C)
Q = 500.0 g × 0.9930 J/g·K × 65.7 K
Q = 32,622.45 J
To convert the result to kilojoules (kJ), we divide by 1000:
Q = 32.62245 kJ
Thus, the amount of heat absorbed by the sample is 32.6 kJ.
To learn more about the specific heat, follow the link:
https://brainly.com/question/11297584
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