Question

A 1.00-L insulated bottle is full of tea at 92.0°C. You pour out a mug of tea and immediately s remaining in the bottle that results from the admission of air at room t screw the stopper back on the bottle. Find the change in temperature of the tea (Let the room temperature be 20.0°C and assume that you poured out 185 cm2 of tea. Take the malar mass of a r as 28.9 g malandpar-1.20×10 3 g/cm, Here we define a "mo natomic ideal gas" to have molar specific heats C-2R and C,"SR, and a diatomic ideal gas" to have Cv÷and CP-7R.) 0 00431x Your response differs from the correct answer by more than 10%. Double check your calculations-c

Answer 1

Consider (800-185) cm³ of tea at 90.0^o C mixing with 200 cm³ of diatomic ideal gas at 20.0^o C.

Q_cold = -Q_not

or   

ma, CP.air(Tf _ Tiair)--In w cul ▲T.c

ー(ρν)arSpar(90.00-20.00) ーrnairCP.air(T「T,.air ) ▲T),,, = =

where we have anticipated that the final temperature of the mixture will be close to 90.0^oC

The molar specific heat of air is

7 CP air 12r

So the specific heat per gram is

1.00mol 28.9g CP.air _ 2 (ΛΊ

and

(1.20 × 10 3g/on3)(200C7113)|(1.01J g. C ) ( 70.0°C,) (1.00g/cm3) (800 185)cm31 (4.186.J/g.oC) ▲T)',

$(\Delta T)_w\approx -6.59 \times 10^{-3}^{\circ}C$

The change of temperature for the water is

between 10^-3oC and 10^-2oC