An ideal Carnot engine operates between reservoirs having temps of 125 degrees C and -20.0 degrees C. Each cycle the heat expelled by this engine is used to melt 30.0 g of ice at 0.00 degrees C. The heat of fusion of water is 3.34 X 10^5 J/kg and the heat of vaporization of water is 2.25 x 10^6 J/kg. How much work is does this engine do each cycle? How much heat per cycle does this engine absorb at the hot reservoir?

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hamzaahmeds hamzaahmeds · Answer 1 · 2020-10-11T02:35:01+02:00

Answer:

Q₁ = 15.78 KJ

W = 5.76 KJ

Explanation:

First we find efficiency of this ideal Carnot engine:

Efficiency = 1 - T₂/T₁

T₁ = source temperature = 125°C + 273 = 398 K

T₂ = Sink Temperature = -20°C + 273 = 253 K

Therefore,

Efficiency = 1 - 253 K/398 K

Efficiency = 0.365 = 36.5%

Now, we find expelled heat (Q₂), by using latent heat of fusion of ice:

Q₂ = mH

where,

m = mass of ice melted = 30 g = 0.03 kg

H = Latent Heat of Fusion of Ice = 3.34 x 10⁵ J/kg

Therefore,

Q₂ = (0.03 kg)(3.34 x 10⁵ J/kg)

Q₂ = 1.002 x 10⁴ J = 10.02 KJ

Now, we use another formula of efficiency to find absorbed heat (Q₁):

Efficiency = 1 - Q₂/Q₁

0.365 = 1 - 10.02 KJ/Q₁

10.02 KJ/Q₁ = 1 - 0.365

Q₁ = 10.02 KJ/0.635

Q₁ = 15.78 KJ

Now, for work done we have the formula:

W = Q₁ - Q₂

W = 15.78 KJ - 10.02 KJ

W = 5.76 KJ