13. (a) We refer to the copper block as block 1 and the lead block as block 2. The equilibrium temperatureTfsatisFesm1c1(Tf−Ti,1)+m2c2(Tf−Ti,2) = 0, which we solve forTf:Tf=m1c1Ti,1+m2c2Ti,2m1c1+m2c2=(50 g)(386 J/kg·K)(400 K) + (100 g)(128 J/kg·K)(200 K)(50 g)(386 J/kg·K) + (100 g)(128 J/kg·K)=320 K.(b) Since the two-block system in thermally insulated from the environment, the change in internal
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