A 1.00-L insulated bottle is full of tea at 88.0°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 160 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 = 3/2R and CP = 5/2 R, and a "diatomic ideal gas" to have CV = 5/2 R and CP = 7/2 R.)

Question

A 1.00-L insulated bottle is full of tea at 88.0°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 160 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 = 3/2R and CP = 5/2 R, and a "diatomic ideal gas" to have CV = 5/2 R and CP = 7/2 R.)

Answer 1

how do you got 1.2?

can someone explain?

Answer 2

To solve this problem, we need to consider the heat transfer between the tea and the air, assuming that it occurs at a constant pressure.

Given:
- Initial temperature of the tea (T1) = 88.0°C
- Final temperature of the tea after the admission of air (T2) = ?
- Initial volume of tea (V1) = 1.00 L
- Volume of tea poured out (Vp) = 160 cm³
- Initial temperature of air (Tair) = 20.0°C
- Molar mass of air (Mair) = 28.9 g/mol
- Density of air (ρair) = 1.20 x 10⁻³ g/cm³

We first need to calculate the mass of the air that entered the bottle:

Mass of air (m air) = density x volume
mair = ρair x Vp

Converting Vp to liters:
Vp = 160 cm³ ≈ 0.160 L

Substituting the values:
mair = (1.20 x 10⁻³ g/cm³) x (0.160 L)
mair = 0.192 g

Next, we calculate the change in heat (ΔQ) by using the specific heat capacity (Cp) at constant pressure:

ΔQ = m x Cp x ΔT

Since the tea and air are assumed to reach the same final temperature, we can set up the following equation:

(Initial heat of tea + Heat gained by the air) = Final heat of tea

Considering that there is no heat transferred to the surroundings:

mtea x Ctea x ΔTtea + mair x Cair x ΔTair = mtea x Ctea x ΔTfinal

Here, we need to consider whether the diatomic or monatomic ideal gas model should be used for air. According to the molar specific heat capacities mentioned in the question:

For monatomic ideal gas:
Cp = (5/2)R
Cv = (3/2)R

For diatomic ideal gas:
Cp = (7/2)R
Cv = (5/2)R

Since air is a diatomic gas, we will use diatomic Cp for air and assume Cv = Cp - R.

R = 8.314 J/(mol·K) (Universal gas constant)

Cp = (7/2)R
Cv = (5/2)R

Now, we can calculate the change in heat:

mtea x Ctea x ΔTtea + mair x Cair x ΔTair = mtea x Ctea x ΔTfinal

Substituting the appropriate values:

(1.0 kg) x Ctea x (88.0 - 20.0) + (0.192 kg) x Cair x (88.0 - 20.0) = (1.0 kg) x Ctea x (ΔTfinal)

Simplifying,

(88.0 - 20.0) [Ctea + (0.192 kg) x Cair] = Ctea x ΔTfinal

(ΔTfinal) = (88.0 - 20.0) [Ctea + (0.192 kg) x Cair] / Ctea

Finally, substitute the appropriate specific heat capacities:

Ctea = (Cp)tea - R
Cair = (Cp)air - R

(ΔTfinal) = (88.0 - 20.0) [(Cp)tea - R + (0.192 kg) x ((Cp)air - R)] / [(Cp)tea - R]

Answer 3

To find the change in temperature of the tea remaining in the bottle, we need to calculate the heat transfer between the tea and the air that was admitted into the bottle.

First, let's calculate the mass of air that entered the bottle. We are given the volume of tea poured out, which is 160 cm³. Since the density of air is 1.20 × 10^(-3) g/cm³, we can calculate the mass of the air using the formula:

mass_air = volume * density = 160 cm³ * 1.20 × 10^(-3) g/cm³

Next, we need to calculate the heat transferred from the tea to the air. The heat transfer can be calculated using the equation:

Q = m * C * ΔT

Where:
Q is the heat transfer,
m is the mass of the substance,
C is the specific heat capacity of the substance, and
ΔT is the change in temperature.

In this case, the specific heat capacity of the air depends on whether it is considered a monatomic or diatomic ideal gas. We can obtain the molar specific heat capacities for the air from the given information.

Now, let's calculate the molar mass of air:

Molar mass_air = 28.9 g/mol

Since we know the molar mass of air, we can calculate the number of moles of air using the formula:

moles_air = mass_air / molar mass_air

Next, we need to determine whether air is a monatomic or diatomic ideal gas. This depends on the specific heat ratio (γ), which is the ratio of the specific heat capacities of air at constant pressure (CP) and constant volume (CV).

To calculate γ, we can use the formula:

γ = CP / CV

For a monatomic ideal gas, γ is equal to 5/3. For a diatomic ideal gas, γ is equal to 7/5.

Once we determine the type of ideal gas, we can calculate the specific heat capacities using the known molar specific heat capacities. If air is monatomic, we use CV = (3/2)R and CP = (5/2)R. If air is diatomic, we use CV = (5/2)R and CP = (7/2)R, where R is the ideal gas constant (R = 8.314 J/(mol·K)).

Now we have all the necessary information to calculate the change in temperature:

ΔT = Q / (m * C)

where Q is the heat transfer, m is the mass of the tea, and C is the specific heat capacity of tea.

By substituting the values into the equation and solving, you can find the change in temperature of the tea remaining in the bottle.