1. The problem statement, all variables and given/known data Thorium (with half-life T1/2 = 1.913 yr. and atomic mass 228.028715 u) undergoes alpha decay and produces radium (atomic mass 224.020186 u) as a daughter nucleus. (Assume the alpha particle has atomic mass 4.002603 u.) (a) What is the decay constant of thorium? (Note that the answer must be in units of hrs-1.) -- I got this correct (4.14 * 10^-5) (c) The energy released from the decay of 10 g of thorium is used to heat 3.8 kg of water (assume all the energy released goes into the water). What is the change in temperature of the water after 1 hr.? --------HELP: This is a multi-step problem. Among the things you must do are: i) determine the heat given off by the decay of one thorium atom, ii) determine the number of thorium atoms that decay to get the total heat given off, iii) determine the temperature change of the water knowing how much heat is given off. HELP: To determine the heat given off, figure out the difference in mass of the reactants and products and use E = mc2. Be careful of your units. HELP: To determine the amount of thorium that decays, figure out how many thorium atoms there were initially and then determine how many there are after 1 hrs. As always, watch your units. HELP: Once you know the heat released, use the heat capacity of water (4186 J/kg-K) and the mass of water to find the temperature change. 2. Relevant equations N = NOe^-λt C= mcT E= mc2 3. The attempt at a solution - 10g(1/228.028715)(6.022 * 10^23) = 2.64 * 10^22 atoms - N = NOe^-λt = (2.64*10^22)e^-(4.14 * 10^-5)(1) = 2.63989 * 10^22 atoms -- From this point I'm having problems figuring out the difference in mass between initial and final I think. Should the initial mass be: (2.64 * 10^22)(1 mo/ (6.022 * 10^23)(228.028715 g/ 1 mol) ...? What should the final mass be? I know from that point you plug in the change in mass into E = mc2, and take that energy and put it in for Q in Q=mcT (with the mass here being the mass of the water), and solve for T. The T I get here - 273 would be the degrees in celsius. Help!