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natita [175]
3 years ago
6

An electric power plant uses energy from burning coal to generate steam at 450∘C. The plant is cooled by 20∘C water from a nearb

y river. If burning coal provides 100 MJ of heat, what is the theoretical minimum amount of heat that must be transferred to the river during the conversion of heat to electric energy? A. 100 MJ. B. 90 MJ. C. 60 MJ. D. 40 MJ.
Physics
1 answer:
Goryan [66]3 years ago
6 0

Answer:

40 MJ (D)

Explanation:

Quantity of heat (Qh) = 100 MJ

temperature of steam (Th) = 450°c = 450 + 273 = 723 K

emperature of water (TI) = 20 °c = 20 + 273 = 293 k

efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

 \frac{100x10x^{6}-Qi }{100x10^{6}} = \frac{723-293}{723}

100x10^{6} - Qi= 0.5947 x 100 x 10 ^{6}

100x10^{6} - (0.5947 x 100x10^{6}) = Qi

Qi = 40.5 MJ equivalent to 40 MJ (D)

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By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

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