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The fuel released 90 calories of heat.
Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:
(1)
Where:
- Mass of the sample, in grams.
- Specific heat of water, in calories per gram-degree Celsius.
- Temperature change, in degrees Celsius.
If we know that ,
and
, then the heat released by the fuel is:
The fuel released 90 calories of heat.
We kindly invite to check this question on sensible heat: brainly.com/question/11325154
Answer:
(a) Δ = 2.881 J/K; Δ
= -2.881 J/K; total change in entropy = 0
(b)Δ = 2.881 J/K; Δ
= 0 ; total change in entropy = 2.881 J/K
(c) Δ = 0 ; Δ
= 0 ; total change in entropy = 0
Explanation:
In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:
mass (m) = 14 g
Temperature = 298 K
Pressure = 1.00 bar
Initial volume =
Final volume = =
(a) Change in entropy of the system Δ =
where R = 8.314 J/(mol*K)
n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles
Δ = 0.5*8.314*ln2 = 2.881 J/K
Change in entropy of the surrounding Δ = -2.881 J/K
Therefore, for a reversible process, the total change in entropy = Δ+Δ
= 2.881 - 2.881 = 0
(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ = 2.881 J/K
Since surrounding does not change in this process Δ = 0.
total change in entropy = Δ+Δ
= 2.881 - 0 = 2.88 J/K
(c) For an adiabatic reversible expansion, q(rev) = 0, thus:
Δ = 0
Since heat energy is not transferred from the system to the surrounding
Δ = 0
total change in entropy = Δ+Δ
= 0