A 45.4 grams of a gold metal sample was heated with the amount of energy of 228 J. Which of the following would the most plausible temperature change of the gold sample? The specific heat of gold is 0.131 J/g.oC.
12.8 oC to 38.3 oC
24.4 oC to 81.6 oC
23.8 oC to 62.1 oC
38.3 oC to 52.1 oC
q = mass Au x specific heat Au x delta T.
228 = 45.3 x 0.131 x delta T.
Solve for delta T, the check the answers and see which matches delta.
To find the temperature change of the gold sample, we can use the formula:
q = m * c * ΔT
Where:
q - energy absorbed or released by the substance (in this case, 228 J)
m - mass of the substance (gold sample) (45.4 g)
c - specific heat capacity of the substance (0.131 J/g.oC)
ΔT - change in temperature
Rearranging the formula, we have:
ΔT = q / (m * c)
Plugging in the given values, we can calculate the temperature change for each option:
For option 1 (12.8 oC to 38.3 oC):
ΔT = 228 J / (45.4 g * 0.131 J/g.oC)
ΔT ≈ 38.5 oC
For option 2 (24.4 oC to 81.6 oC):
ΔT = 228 J / (45.4 g * 0.131 J/g.oC)
ΔT ≈ 80.7 oC
For option 3 (23.8 oC to 62.1 oC):
ΔT = 228 J / (45.4 g * 0.131 J/g.oC)
ΔT ≈ 60.5 oC
For option 4 (38.3 oC to 52.1 oC):
ΔT = 228 J / (45.4 g * 0.131 J/g.oC)
ΔT ≈ 14.9 oC
After calculations, we find that the most plausible temperature change of the gold sample is from 24.4 oC to 81.6 oC (option 2).