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Lera25 [3.4K]
3 years ago
13

For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 175-cm-long vertical cylinder with both the top a

nd bottom surfaces insulated and with the side surface at an average temperature of 35 . If the rate of heat loss from this man to the environment at is 336 W, determine the rate of entropy transfer from the body of this person accompanying heat transfer, in W/K.
Chemistry
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

Sgen = 0.0366 W/K

Explanation:

for the body:

∴ Q = - 336 W...rate of heat loss

∴ T surface = 35°C ≅ 308 K

the rate of entropy transfer from the body:

⇒ ΔS = - Q/Ts

for the enviroment:

⇒ ΔS = Q/Te

∴ assuming: T = Tenv = 25°C ≅ 298 K

resulting in a net variation in the universe:

⇒ Sgen = ΔS = Q/Tenviroment - Q/Tsurface = Q(Ts - Te)/Ts*Te

⇒ Sgen = (336( 308-298))/(308×298) = 3360 WK/91784 K² = 0.0366 W/K

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What fraction of a Sr-90 sample remains unchanged after 87.3 years
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The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span>t_{1/2} = 28.8 years

We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{87.3 years}{28.8 years}
⇒ n=3.03
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</span>
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