The graph above shows the changes in temperature recorded for the 2.00 l of h2o surrounding a constant-volume container in which
a 1.00 g sample of benzoic acid was combusted. Assume that heat was not absorbed by the container or lost to the surroundings
1 answer:
Answer:
25.2 kJ
Explanation:
The complete question is presented in the attached image to this answer.
Note that, the heat gained by the 2.00 L of water to raise its temperature from the initial value to its final value comes entirely from the combustion of the benzoic acid since there are no heat losses to the containing vessel or to the environment.
So, to obtained the heat released from the combustion of benzoic acid, we just calculate the heat required to raise the temperature of the water.
Q = mCΔT
To calculate the mass of water,
Density = (mass)/(volume)
Mass = Density × volume
Density = 1 g/mL
Volume = 2.00 L = 2000 mL
Mass = 1 × 2000 = 2000 g
C = specific heat capacity of water = 4.2 J/g.°C
ΔT = (final temperature) - (Initial temperature)
From the graph,
Final temperature of water = 25°C
Initial temperature of water = 22°C
ΔT = 25 - 22 = 3°C
Q = (2000×4.2×3) = 25,200 J = 25.2 kJ
Hope this Helps!!!
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Answer:
The temperature difference of the body after 3 hours = 5.16 K
Explanation:
we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹
or, 1.2 mole.h⁻¹
The average heat evolved by the oxidation of foodstuffs is then:
⇒ Q avg = = 7.2 kj.h⁻¹.Kg⁻¹
the heat produced after 3 h would be:
= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h
= 21.6 kj. kg⁻¹
= 21.6 x 10³ j kg⁻¹
We know Qp = Cp x ΔT
Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹
⇒ ΔT =
⇒ ΔT =
⇒ ΔT = 5.16 K