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lolorichards123
04.11.2021 •
Chemistry
A 20.0 g sample of aluminum (specific heat = 0.902) g-1 oC-1) with an initial temperature of 2.5°C is heated with 427 J of
energy. What is the final temperature of the sample?
O 72.3°C
O 23.7°C
0 26.2°C
0 74.8°C
O 24.9°C
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Ответ:
T(final) = 26.2°C (3 sig. figs.)
Explanation:
Q = m·c·ΔT
m = mass of sample of interest = 20.0g
c = specific heat of sample of interest = 0.902j·g⁻¹·C⁻¹
ΔT = Temperature change = T(final) - T(initial) = T(f) - 2.5°C
Q = 427 joules
Q = m·c·ΔT => = ΔT = Q/m·c => T(f) = Q/m·c + T(i) = (427j/20.0g·0.902j·g⁻¹·C⁻¹) + 2.5°C = 26.16962306°C (calc. ans.) ≅ 26.2°C (3 sig. figs.)
Ответ:
A line is an undefined term because it : Contains infinite number of points
In Geometry, unless it's stated, a line will extend in one dimension and goes on forever in both ways
Step-by-step explanation: