Question

A 1.00-L insulated bottle is full of tea at 92.5°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 220 cm3 of tea. Take the molar mass of air as 28.9 g/mol and Pair 1.20 x 103 o/cm2. Here we define a imonatomic ideal gas' to have mola speic heas Cy-^R and C, -32, and a "diatomic ideal gas to have 7 C,, = R and Ca-LA.) V 2

Answer 1

Solution: When tea is poured out a mug of tea T (220 cm3), the cool air at 20°C immediately Tea Air rushes into the bottle and fill the vacant space (vacuum). Since there is no heat exchange to the environment so the air in contact to the hot tea heats up adiabatically. The heat supplied by water to air is equal to the work done by water that change the temperature of the air. Let T, be the final or equilibrium temperature of tea+ air system Now the volume of tea remain in the bottle (1000-220) -780 cm Therefore, mass of tea in the bottle M, 780cm3 x1gm/cm3 780 gm 0.78 kg (we take density of tealgm/cm. The = specific heat of teas 4200 J /kg °C The heat supplied by the tea to cool air H 0.78x4200(92.5-T)J 3276(92.5- T. The work done on cool air (in adiabatic change) W (T,-20), R y-1 8.31 J/ (mole-K), u mole 7 5 1.4 (air diatomic gas) 2 2 of the air, y the specific heat of the air, y C, /C Mass of the air in the bottle m = pV = 1.20x10 x220 = 0.264 gm. So number of moles of air 28.9 in the bottle u = =109.47 0.264 109.47 x8.31 (T,-20) 2274.24 (T,-20)J (T, - 20) Y-1 Therefore, W = 1.4-1 = W or 3276x(92.5-T, )J 2274.24x (T, - 20)J For equilibrium of temperature H or (2274.24+3276) T = 3276 x 92.5 2274.24x20 T, 62.79°C Thus (2274.24 3276)T, = 3276 x 92.5 2274.24 x 20