Question

The half-life for the alpha decay of ²³⁸Pu to form ²³⁴U is 87.7 y. Suppose the energy of the alpha particle is 6.01 MeV and that 1.18 g of ²³⁸Pu, freed of any other radioactive materials, were placed in a calorimeter containing 48.1 g of water, initially at 22.0°C. Calculate the temperature the water would reach after 1.07 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 ²³⁸Pu atom to be 238.050 u. (1 eV = 1.6022 × 10^-19 J)

°C

Answer 1

Pu-238 -> ( alpha decay) U-234

T_1/2 = 87.7 Yrs

decay const. = 0.693/87.7*365*24 = 6.25E-7 /hr

mass of Pu-238 = 1.18 gm

Initial number of Pu-238 atoms N_o = 1.18/238 * 6.02E+23 = 2.985 E+21

Half life is much larger than the time period 1.07 hr, the decay rate of Pu-238 can be taken as const through the time interval of 1.07 hrs

decay rate = N_o /hr

number of alpha decays during the period N = 2.985E+21*6.25E-7 *1.07

= 1.997E+15

For each alpha 6.01 Mev of energy is released

total energy released = 6.01*1.997E+15 Mev

= 6.01*1.997E+15 * 1.602E-13 J

= 1922.7 J

mass of water = 48.1 gm

Initial temp = 22^o C

final temp T^oC

specific heat of water = 4.184 J/K-gm

48.1( T-22) *4.184 = 1922.7

T = 31.55^o C