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A liquid at 0^@ C is poured in a glass b...

A liquid at `0^@ C` is poured in a glass beaker of volume 600 `cm^3` to fill it completely. The beaker is then heated to `90^@ `C. How much liquid will overflow?
` ( gamma_(" liquid") = 1.75 xx 10^(-4) //^@ , gamma _("glass")= 2.75 xx 10^(-5) //^@C )`

Text Solution

Verified by Experts

Given
` V_1 = 600 cm^3`
` T_1 = 0^@ C `
`T_2 = 90^@ C`
we have ` gamma = (V_2 - V_1)/( V_1 (T_2 -T_1))`
we have ` gamma = (V_2 - V_1)/( V_1 (T_2 -T_1))`
` therefore ` increase is volume `=V_2 - V_1 = gamma V_1 (T_2 -T_1)`
Increase in volume of beaker
` = gamma_("glass") xx V_1 (T_2 -T_1)`
`= 2.75 xx 10^(-5) xx 600 xx (90-0)`
` = 2.75 xx 10^(-5) xx 600 xx 90`
` = 148 500 xx 10^(-5) cm^3`
` therefore ` increase in volume of breaker `= 1.485 cm^3`
increase in volume of liquid
`= gamma_("liquid") xx V_1 (T_2 -T_1)`
`= 1.75 xx 10^(-4) xx 600 xx (90-0)`
`= 1.75 xx 10^(-4) xx 600 xx 90`
` = 94500 xx 10^(-4) cm^3`
` therefore `increase in volume of liquid `= 9.45 cm^3`
` therefore ` volume of liquid which overflows
`=(9.45 -1.485 ) cm^3`
`=7.965 cm^3`
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