Heat is transferred at a rate of 2 kW from a hot reservoir at 825 K to a cold reservoir at 300 K. Calculate the rate at which the entropy of the two reservoirs changes. (Round the final answer to six decimal places.) The rate at which the entropy of the two reservoirs changes is 00424 kW/K.

0.004243 KW/K.

Explanation:

From the question given above, the following data were obtained:

Heat transfered (Q) = 2 KW

Temperature of hot reservoir (Tₕ) = 825 K

Temperature of cold reservoir (T꜀) = 300 K

Next, we shall determine the entropy of both reservoir. This can be obtained as follow:

For hot reservoir:

Heat transfered (Q) = 2 KW

Temperature of hot reservoir (Tₕ) = 825 K

Entropy of hot reservoir (Sₕ) =?

Sₕ = Q/Tₕ

Sₕ = 2/825

Sₕ = 0.002424 KW/K

For cold reservoir:

Heat transfered (Q) = 2 KW

Temperature of cold reservoir (T꜀) = 300 K

Entropy of cold reservoir (S꜀) =?

S꜀ = Q/T꜀

S꜀ = 2/300

S꜀ = 0.006667 KW/K

Finally, we shall determine the change in the entropy of the two reservoirs. This can be obtained as follow:

Entropy of hot reservoir (Sₕ) = 0.002424 KW/K

Entropy of cold reservoir (S꜀) = 0.006667 KW/K

Change in entropy (ΔS) =?

ΔS = S꜀ – Sₕ

ΔS = 0.006667 – 0.002424

ΔS = 0.004243 KW/K.

C. hazardous waste

Explanation:

The given picture depicts the sign for hazardous waste over the bin.

Hazardous waste is the chemical waste that can be either in the form of gases, solid or liquid is capable of causing damage to the organisms. The waste can be corrosive, reactive, ignitable, toxic and effect of the waste may remain for long and drastic. A hazardous waste should be available in a laboratory so as to prevent accidental fire and explosion, burns and injuries.