Question

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 JOC and water 4.18 Jig C O 308 K O 325K 348 K Loooo x x

Answer 1

I Page Page solution i Let the final temperature of the water be It mass of iron, m; = 25.09 initial temperature, To=398K: tinal temperature= T which need to be determine. into note i Water both after dipping the iron piece , the final temperature of the iron & water will be same. specific heat capacity of iron, Cpliron) =0.449 I/.°C. initial temperature of Water F298k. 50.449 J/g. K. initial temperature of iron inisial temperature _ a we & Iron will loss heat (9) .. Giron = m Cp (iron) AT - = mcp X (Tx-T;) Piton 3 254 X 09949 FK *(1-3982) ESTIMA 9;yon = 11:225 ( 39 iron will release heat. mass of water, mw = volume of water X density = 25 ml x I gm/m1 mw = 25gm .

Page Page Y initial temperature of water, Ti = 298K. forul temperature of water, = Tt. heat capacity of water, Cp (H20 (4) = 4:185/9.0C = 4'18 J/g. K initial temperature of water < Initial temp of iron. i warter will absorbed heat (Aw) .. = ชาพ X CH.9 () X 4T = Mwt Cp (H20 (1)) x (It-Ti) = 25 qm x 4 18 7/8 R X (Tz - 298K) = 104.5 (It-298K) J/K be by iron will pabsorbed 28K now, heat released water , i qw = -Tiron a) Toy's ( I-398K) = -11225(T6 – 398K) Je HIHIHUME => T04'5 T4 — 31141 K = - 11.225 TE +446755 - 115'725 T = 35608:55 K. => TI = 307.7 K. Elnal temperature of water is 3077K

part-2 2 H2O2(l) a mol → 2H2048) + O2(8) AH = - 196 KJ. Amol Imo) N mass of hydrogen peroxide (H₂O2) = 2.00g. Molecular mass of H₂O2 = 2XH+ 2XO = 2x1 + 2 x 16, = 34 gm /mol. i. mol of H₂O = given muss Molecalon mass INTSULLA . = = zy home moli from the reaction ; For a mol H₂²2 reaction enthalpy is -196 kJ For, I mol H₂O₂ a n is =-116 KJ for, 1 mol H2²2 reaction on halpy iz=- 19 x 7 KJ | - =-5.76 kJ For 1 mol H2O2,9=-5.76 kJ me