A 1.00−L insulated bottle is full of tea at 94.5∘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 195 cm3 of tea. Take the molar mass of air as 28.9 g/mol and ρair = 1.20×10−3 g/cm3. Here we define a "monatomic ideal gas" to have molar specific heats CV = 32 R and CP = 52 R, and a "diatomic ideal gas" to have CV = 52 R and CP = 72 R.) ∘C

A 1.00−L insulated bottle is full of tea at 94.5∘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 195 cm3 of tea. Take the molar mass of air as 28.9 g/mol and ρair = 1.20×10−3 g/cm3. Here we define a "monatomic ideal gas" to have molar specific heats CV = 32 R and CP = 52 R, and a "diatomic ideal gas" to have CV = 52 R and CP = 72 R.) ∘C

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A 1.00 L insulated bottle is full of tea at 94.5 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 195 c m 3 of tea. Take the molar mass of air as 28.9 g / m o l and ρ air = 1.20 × 10 3 g / c m 3 . Here we define a "monatomic ideal gas" to have molar specific heats C V = 3 2 R and C P = 5 2 R , and a "diatomic ideal gas" to have C V = 5 2 R and C P = 7 2 R . ) C

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