Question

A 1.00-L insulated bottle is full of tea at 90.5°C. You pour out a mug of tea and immediately screw the stopper back on the bottle. Find the change in temperature of the tea remaining in the bottle that results from the admission of air at room temperature. (Let the room temperature be 20.0°C and assume that you poured out 205 cm3 of tea. Take the molar mass of air as 28.9 g/mol and ραǐr-1.20 x 10-3 g/cm3. Here we define a "monatomic ideal gas" to have molar specific heats C,--R and CP-ーR, and a "diatomic ideal gas" to have CV--R and C,--R.) 0.052 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.°C Need Help? Read It

Answer 1

OUD-2OS hea aft paring li mug 3 -6 =795 メ10 3 So u vdunmt aiv i'w bote 205 2 . aosri 7 - 3 3 メ10 9 5710 So, iu mans f ha in ilu bruu -3 -3 x 2 メ10 *10°×10 a05ID . A ans -b ;246メ16 uninu hcat ef tea St T/k c ニイ86 。 NBw, -3 mde. no. om sf air : 24610 8.51 XCD 3 28-9メLO gei A ar Atx ts main diatomi N2 0) Ju s mdor epeific &ear a Cont wlume ( ne sanen ^ Kept in Comt vdmme )

hen 7 3 83/x (t-o) 791メ16 メ 486x Oo'5-t): 3 DI17S 8 33 28.0St d6495 90.4962 Chanrpe in demp fr tea= (90.5-90-496= O O038