Question 54: Thermodynamics & KTG

Question

A glass of hot water cools from $$90^{\circ} \mathrm{C}$$ to $$70^{\circ} \mathrm{C}$$ in 3 minutes when the temperature of surroundings is $$20^{\circ} \mathrm{C}$$. What is the time taken by the glass of hot water to cool from $$60^{\circ} \mathrm{C}$$ to $$40^{\circ} \mathrm{C}$$ if the surrounding temperature remains the same at $$20^{\circ} \mathrm{C}$$ ?

Accepted Answer

To determine the total mass of the gas in the box, we first need to understand the given problem and apply the principles of kinetic theory of gases. Let's summarize the given data:
Side length of the cubical box, $$L = 1 \ 	ext{m}$$
Pressure of the gas, $$P = 100 \ 	ext{Nm}^{-2}$$
Observation time, $$t = 1 \ 	ext{s}$$
Number of hits on the wall, $$n = 500$$
We are given that the atom travels with the root mean square (rms) speed (denoted by $$v_{rms}$$) parallel to one edge of the cube, and makes 500 hits on the wall in 1 second. From this, we can find the $$v_{rms}$$ as follows:
Each hit corresponds to the atom travelling across the side of the cube and bouncing back. The distance travelled between two consecutive hits is $$2L$$, as the atom goes to the wall and returns. Therefore, the total distance travelled in 1 second is:
$$ 500 	imes 2L = 500 	imes 2 	imes 1 = 1000 \ 	ext{m} $$
So, the rms speed $$v_{rms}$$ is:
$$ v_{rms} = \frac{1000 \ 	ext{m}}{1 \ 	ext{s}} = 1000 \ 	ext{m/s} $$
Now, using the formula from the kinetic theory of gases for pressure:
$$ P = \frac{1}{3} ho v_{rms}^2 $$
where $$ho$$ is the density of the gas. We can express density $$ho$$ as:
$$ ho = \frac{m}{V} $$
Here, $$m$$ is the mass of the gas and $$V$$ is the volume of the box. For a cubical box, $$V = L^3$$, so:
$$ V = 1^3 \ 	ext{m}^3 = 1 \ 	ext{m}^3 $$
Substituting the values into the pressure equation:
$$ 100 \ 	ext{Nm}^{-2} = \frac{1}{3} \left( \frac{m}{1} ight) (1000)^2 $$
Simplifying, we get:
$$ 100 = \frac{1}{3} m 	imes 1000000 $$
$$ 100 = \frac{1000000}{3} m $$
$$ m = \frac{100 	imes 3}{1000000} $$
$$ m = 0.0003 \ 	ext{kg} $$
Converting this mass to grams:
$$ m = 0.0003 \ 	ext{kg} 	imes 1000 \ 	ext{g/kg} = 0.3 \ 	ext{g} $$
Therefore, the total mass of the gas in the box is:
Option B: 0.3 grams

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Thermodynamics & KTG

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