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allochka39001 [22]

2 years ago

14

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask

is filled, the glycerin cools and the flask warms so that the combination of glycerin and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C
What is the volume of the glycerin when it cools to 20.0°C?

1 answer:

Charra [1.4K]2 years ago

7 0

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

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Therefore the required solution is

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$s^0(T1)$ = 1.8279 kJ/kg.K

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