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odalysesquermon
12.04.2020 •
Chemistry
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C when it absorbs 3.42 kilojoules of energy.
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Ответ:
0.036 J/g°C
Explanation:
The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Given that,
Q = 3.42 Kilojoules
[Convert 3.42 kilojoules to joules
If 1 kilojoule = 1000 joules
3.42 kilojoules = 3.42 x 1000 = 3420J]
Mass = 2.508Kg
[Convert 2.508 kg to grams
If 1 kg = 1000 grams
2.508kg = 2.508 x 1000 = 2508g]
C = ? (let unknown value be Z)
Φ = (Final temperature - Initial temperature)
= 42.061°C - 4.051°C
= 38.01°C
Apply the formula, Q = MCΦ
3420J = 2508g x Z x 38.01°C
3420J = 95329.08g•°C x Z
Z = (3420J / 95329.08g•°C)
Z = 0.03588 J/g°C
Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C
Thus, the specific heat of the substance is 0.036 J/g°C
Ответ:
c) carbon and boron
Explanation:
They have the same number of neutrons but different number of protons.
They both contain 7 neutrons