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Naya [18.7K]
3 years ago
7

Q4: The average heat evolved by the oxidation of food in an average adult per hour per kilogram of body weight is 7.20 ୩୎ ୩୥ ୦୰.

Assume the weight of an average adult is 62.0 kkg. Suppose the total heat evolved by this oxidation is tranderred into the surroundings over a period of 1 week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at 293K.
Chemistry
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

The entropy change of the surroundings associated with this heat transfer is 255.9J/K

Explanation:

The heat evolved by the oxidation of food is 7.2J/Kg.h thus, multiplying the heat by the weight of the person and by the time the total heat (q) is obtained. In this case, the weight is 62.0Kg and a weak has 168 hours.

q = 7.2\frac{J}{Kg.h}x62.0Kgx168h = 74995.2J

The entropy change of the surrounding associated with this heat transfer can be calculated by the ratio between the heat exchanged and the temperature.

ΔS = \frac{q}{T} = \frac{74995.2J}{293K} = 255.9J/K

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2 years ago
What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor
Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

CH_3COOH(l)\rightarrow CH_3COOH(g)..[1]

Heat of vaporization of acetic acid = H^o_{vap}=24.3 kJ/mol

CH_3COOH(s)\rightarrow CH_3COOH(l)..[2]

Heat of fusion of acetic acid = H^o_{fus}=10.8 kJ/mol

Heat of sublimation of acetic acid = H^o_{sub}=?

CH_3COOH(s)\rightarrow CH_3COOH(g)..[3]

[1] + [2] = [3] (Hess's law)

H^o_{sub}=H^o_{vap}+H^o_{fus}

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35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0
2 years ago
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
2 years ago
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