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2020-06-13 12:40 am
Use the heat equation to calculate the energy, in joules and calories, for each of the following (see the table):
回答 (2)
2020-06-13 1:06 am
Part E
Energy lost
= m c ΔT
= (15.2 g) × (2.46 J/g°C) × {[52.5 - (-42.0)] °C}
= 15.2 × 2.46 × 94.5 J
= 3530 J
====
Part E
Energy lost
= m c ΔT
= (15.2 g) × (0.588 cal/g°C) × {[52.5 - (-42.0)] °C}
= 15.2 × 0.588 × 94.5 cal
= 845 cal
Energy lost
= m c ΔT
= (15.2 g) × (2.46 J/g°C) × {[52.5 - (-42.0)] °C}
= 15.2 × 2.46 × 94.5 J
= 3530 J
====
Part E
Energy lost
= m c ΔT
= (15.2 g) × (0.588 cal/g°C) × {[52.5 - (-42.0)] °C}
= 15.2 × 0.588 × 94.5 cal
= 845 cal
2020-06-13 1:03 am
You will use the same equation for both Part E and Part F:
q = m c (T2-T1)
q = 15.2 g (0.588 Cal/gC) (-42.0 - 52.5 C) = -845 Cal
The negative sign indicates that heat is lost.
For part F, just use 2.46 J/gC for "c" in the equation.
q = m c (T2-T1)
q = 15.2 g (0.588 Cal/gC) (-42.0 - 52.5 C) = -845 Cal
The negative sign indicates that heat is lost.
For part F, just use 2.46 J/gC for "c" in the equation.
收錄日期: 2021-04-18 18:35:10
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