Question

Decreto 378-4124,248t 9.5.gly Real mabomb a can cals.ol kugle 143 318 ic J heat given oft Ti=24.gol Gz = 88 3 3 TE? 8813 9- cat (5,01)(x - 24.88 Tel Ti 453 kJ of heat is given off when a reaction is carried out in a bomb calorimeter with C = 6.35 kJ/°C. What is the final temperature of the water in the calorimeter if the initial water temperature is 22.4°C ? A. 93.7°C B. 71.3°C C. 48.9°C D. -48.9°C = (8131)(x - 21,0) 467 81314-1761178 To = 17,394 Theat

Answer 1

Since heat is given off by bomb calorimeter. Therefore Q_cal = -453KJ and heat of calorimeter is determined by formula

Q_cal = C_cal * ∆T

In above problem Q_cal and C_cal is given. First we need to find ∆T and then final temperature.

Rearrange the above equation for ∆T.

Therefore,  ∆T= Q_cal / C_cal

∆T = (-453 KJ / 6.35 KJ) * ^oC

∆T = -71.3385 ^oC

since ∆T = T_final - T_initial

Therefore -71.3385 = T_final - 22.4

T_final = -71.3385 + 22.4 = -48.9385 ^oC = -48.9 ^oC

Therefore final temperature of water is -48.9 ^oC.

If you find any mistake, please mention in the comment box.

Thanks.