Question

A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.

Answer 1

Answer:

the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = $9.18*10^{-6} \ m^3$

Increase in volume of the glass =  10⁻³ × 51.00 × $\beta _{glass}$

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = $(9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3$

$8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3$

$\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )$

$\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Thus; the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$