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khushboogarg3877

khushboogarg3877

02.08.2019 •

Physics

When you mix two gases what can you predict about the conservation of volume in this case?

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Ответ:

haydenoo72

15.03.2022

Amagat's law of additive volumes states that we can simply add up the individual volumes of each gas (provided they are at the same temperature and pressure) to get the total volume of the mixture. Conservation of volume is an acceptable assumption for gases (but not always for liquid mixtures). This works for gases since the molecules are very small and only take up a minimal amount of space in a gas.

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Ответ:

MichealM7668

18.05.2023

A. 25°C

B. 0°C/min

C. -0.05/min

D. 66°C

Explanation:

A. In this case, the value of the temperature of coffee will never reach exactly room temperature, but it will approach it. This value to which it approaches is called the limiting value, that is to say, 25°C.

The rate of change of temperature present for this limiting value (25°C) is indeed the limiting value of the rate of cooling. Introducing this value (T=25°C) in the equation:

$\frac{dT}{dt} =k(T-T_{room} )=k(25C-25C)=0$

B. Zero is the limiting value to which the rate of cooling will approach when the coffee temperature reaches the limiting value.

C. To find the constant k in the differential equation, we use the fact that coffee cools at a rate of 2∘C per minute when its temperature is 65∘C.

-2C/min=k(65C-25C)

Note that the sign of -2C/min is negative because the temperature is decreasing with time. Solving for k

$k=\frac{-2C/min}{(65C-25C)} =-0.05/min$

To estimate the temperature of the coffee after 10 minutes, we use Euler's formula. We start with ti=t0=0

$\frac{dT(t=t_{i} )}{dt}=f(t_{i} , T_{i} )$

$T_{i+1} = T_{i}+h*f(t_{i}, T_{i})$

$f(t_{0} , T_{0} )=(-0.05/min)(95C-25C)=-3.5C/min$

$T_{1}(2min) =95C-3.5C/min*2min=88C$

$f(t_1} , T_{1} )=(-0.05/min)(88C-25C)=-3.15C/min$

$T_{2}(4min) =88C-3.15C/min*2min=88C=82C$

$f(t_2} , T_{2} )=(-0.05/min)(82C-25C)=-2.9C/min$

$T_{3}(6min) =82C-2.9C/min*2min=76C$

$f(t_3} , T_{3} )=(-0.05/min)(76C-25C)=-2.6C/min$

$T_{4}(8min) =76C-2.6C/min*2min=71C$

$f(t_{4} , T_{4} )=(-0.05/min)(71C-25C)=-2.3C/min$

$T_{5}(10min) =71C-2.3C/min*2min=66C$

The temperature of the coffee after 10 minutes is 66°C.

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