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TEA [102]
3 years ago
12

In Ray Bradbury's 1953 novel Fahrenheit 451, 451°F is said to be the temperature at which books, which have been banned in the

story, ignite. Convert 451°F to the Celsius scale.
Chemistry
1 answer:
klio [65]3 years ago
8 0

Answer : The temperature in degree Celsius is, 232.78^oC

Explanation :

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=(^oF-32)\times \frac{5}{9}

where,

^oF = temperature in Fahrenheit  

^oC = temperature in centigrade

As we are given the temperature in Fahrenheit is, 451

Now we have to determine the temperature in degree Celsius.

^oC=(^oF-32)\times \frac{5}{9}

^oC=(451^oF-32)\times \frac{5}{9}

^oC=232.78

Therefore, the temperature in degree Celsius is, 232.78^oC

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A brine solution of salt flows at a constant rate of 9 ​L/min into a large tank that initially held 100 L of brine solution in w
Valentin [98]

Answer:

a) C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) The concentration of salt in the tank attains the value of 0.01 kg/L at time, t = 0.0713 min = 4.28s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the Concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F

Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dC/dt = FᵢCᵢ - FC

Fᵢ = 9 L/min, Cᵢ = 0.02 kg/L, F = 9 L/min

dC/dt = 0.18 - 9C

dC/(0.18 - 9C) = dt

∫ dC/(0.18 - 9C) = ∫ dt

(-1/9) In (0.18 - 9C) = t + k

In (0.18 - 9C) = -9t - 9k

-9k = K

In (0.18 - 9C) = K - 9t

At t = 0, C = 0.1/100 = 0.001 kg/L

In (0.18 - 9(0.001)) = K

In 0.171 = K

K = - 1.766

So, the equation describing concentration of salt at anytime in the tank is

In (0.18 - 9C) = -1.766 - 9t

In (0.18 - 9C) = - (9t + 1.766)

0.18 - 9C = e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾

9C = 0.18 - (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) when C = 0.01 kg/L

0.01 = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

0.09 = (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

- (9t + 1.766) = In 0.09

- (9t + 1.766) = -2.408

(9t + 1.766) = 2.408

9t = 2.408 - 1.766 = 0.642

t = 0.642/9 = 0.0713 min = 4.28s

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