Question

The half-life for the alpha decay of ²³²U to form ²²⁸Th is 68.9 y. Suppose the energy of the alpha particle is 5.28 MeV and that 1.09 g of ²³²U, freed of any other radioactive materials, were placed in a calorimeter containing 49.1 g of water, initially at 22.9°C. Calculate the temperature the water would reach after 1.15 hour. Neglect the heat capacity of the calorimeter and heat loss to the surroundings. Take the specific heat of water to be 4.184 J K^-1 g^-1 and the mass of a ²³²U atom to be 232.037 u. (1 eV = 1.6022 × 10^-19 J)

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Answer 1

One alpha particle gives 5.28 MeV

As we know that radioactivity follows 1st order reaction

And writing balanced reaction we get

232U -----> 228Th + 4He(or alpha atom)

half life(t_1/2) = 68.9 y = 68.9 x 365.25 x24 hr = 603977.4 hr

rate constant K = ln2/t_1/2 =0.00000114739 hr^-1

No. of U converted to Th = no. of alpha particles released

lnC₀/C = Kt for 1st order reaction

C₀ = initial mass

C = final mass

ln1.09g/C = 0.00000114739 hr^-1 x 1.15 hr

1.08999856174 g =C

mass change = 1.09-1.08999856174 = 0.00000143825 g

1U atom or 232.037 g gives 1 mole alpha particle

0.00000143825 g gives 6.19840801 x 10^-9 mole alpha particles

1 Mev = 1.60218 x 10^-13 joules

5.28 Mev = 8.45949 x 10^-13 joules/alpha particles

6.19840801 x 10^-9 mole alpha particles = 6.19840801 x 10^-9 x 6.022140857 x 10²³ alpha particles

no. of alpha particles = 3.7327686 x 10¹⁵alpha particles

total energy released = 8.45949 x 10^-13 joules/alpha particles x 3.7327686 x 10¹⁵ alpha particles

= 3157.73187626 joules

heat released is absorbed by water = mCp∆T

49.1g x 4.184 J ⁰C^-1 g-1 x (T2 - 22.9 ⁰C) = 3157.73187626 joules

T2 = 38.2709986071 ⁰C

or T2 = 38.27 ⁰C