Question

1/2 H2(g) + 1/2012(g) - HCl(g) Question 18 (1 point) A piece of iron (mass = 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? The specific heat capacity of iron = 0.449 J/gºC and water = 4.18 J/gºC. 308 K 348 K 325 K 287 K 388 K

Answer 1

Answer: niven, mass of iron mi) = 25g. initial temperature of iron (Ti) = 398k. specific heat capacity of iron (cm) = 0.44g J/gcc. mass of water (un ) = 25g (25me is equi- valent to Initial temperature of T-2988 water Specific heat capacity of water (Cw) = 4.183/9°C let the final temperature be f. Assuming that no heat is lost to the cup con) the surroundings I therefore heat lost by iron will be equal to heat gained by water ie, ; - qm = qw. -(m, & cm x T) = (M2 x CwxDT) where, - qm= heat loss by iron. qw= heat gained by water. DT = Tf - Ti mi, me = mass of iron & water respectively- Cm, Cur= specific heat of iron & Water respectively a

-(m, xcm. (IF - T;)) = (m + Cwx Eff- T;)) => -(259+ 0.449 J/g <> (TR-393)) - (25y +4:18.5/97 +64-7994)) = + 178.702 K 4.18 TF - 1245.64K. -0.4ugTL 4.629 T = 1424-342k. It = 307.7K ANSWER er kan ITA~ 308K
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