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finlep [7]
3 years ago
5

Help help help !!!!!!

Chemistry
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V

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If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

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s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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