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3 years ago

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A graduated cylinder has 32.0 mL of water in it. A 3.0 cm3 rock is dropped into the graduated cylinder. What is the new volume r

eading for the graduated cylinder?

1 answer:

kipiarov [429]3 years ago

4 0

Answer:

The new volume reading for the graduated cylinder is :

<u>35 mL</u>

Explanation:

If a graduated cylinder has 32.0 mL of water in it. A 3.0 cm3 rock is dropped into the graduated cylinder. The new volume reading for the graduated cylinder is :

Here , Volume of rock is given : 3 cm3

<u><em>1. Do unit conversion first</em></u>

$1dm^{3}=1L$ .........(1)

1 dm = 10 cm

$1dm^{3}=(10)^{3}cm^{3}$

$1dm^{3}=1000cm^{3}$ .........(2)

Also.

1 L = 1000 mL........(3)

Insert the value of L and dm3 in equation (1)

$1000cm^{3}=1000mL$

1000 cancel each other

$1cm^{3}=1mL$

So,

$3cm^{3}=3mL$

2. Find the total volume

Volume of rock = 3 mL

Volume of water = 32 mL

New volume =Volume of rock + Volume of water

New volume = 3 mL + 32 mL

<u>New volume = 35 mL</u>

<em>What is need of unit conversion ?</em>

<u><em>Because Only those quantities can be added or subtracted which have same units </em></u>

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You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

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(b) The specific heat of the metal is $0.0944J/g^oC$

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Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

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$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

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$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

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Now put all the given values in the above formula, we get:

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

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c = specific heat of metal = ?

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$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

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