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Agata [3.3K]
4 years ago
9

The half life of cobalt-60 is 5.3 years. If a certain rock currently contains 10.0g of cabals -60 how much cobalt-60 will remain

in the rock after 21.2 years
A. 1.25g

B. 2.50g

C. 0.63g

D. 5.00g
Physics
1 answer:
VLD [36.1K]4 years ago
6 0

T = half life of cobalt-60 = 5.3 years

λ = decay constant for cobalt-60 = ?

decay constant is given as

λ = 0.693/T

inserting the values

λ = 0.693/5.3 = 0.131

N₀ = initial amount of cobalt-60 = 10 g

t = time of decay = 21.2 years

N = final amount of cobalt-60 after time "t"

final amount of cobalt-60 after time "t" is given as

N = N₀ e^{-\lambda t}

inserting the values

N = 10 e^{-(0.131) (21.2)}

N = 10 e^{-(0.131) (21.2)}

N = 10 x 0.0622 g

N = 0.63 g

hence

C. 0.63 g

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
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Answer:

The final equilibrium temperature of the system is T = 12.48^oC

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Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

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Following the change of state of water in the question

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Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

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The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

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Answer with Explanation:

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The Froude number is 1.24611

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