A calorimeter of mass $50 g$ and specific heat capacity $0.42 J g^{- 1} ℃^{- 1}$ contains some mass of water at $20^{o} C$ . A metal piece of mass $20 g$ at $100^{o} C$ is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be $22^{o} C$ . Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece $= 0.3 J g^{- 1} ℃^{- 1}$
specific heat capacity of water $= 4.2 J g^{- 1} ℃^{- 1}$ ]

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Solution

Heat energy given by metal piece $= m \cdot c \cdot Δ T_{1}$
$= 20 \times 0.3 \times (100 - 22)$
$= 468 J$
Heat energy gained by water $= m_{w} \times c_{w} \times Δ T_{2}$
$= m_{w} \times 4.2 \times (22 - 20)$
$= m_{w} \times 8.4 J$
Heat energy gained by calorimeter $= m_{c} \times c_{c} \times Δ T_{2}$
$= 50 \times 0.42 \times (22 - 20)$
$= 42 J$
By principle of calorimeter:
Heat lost $=$ Heat gained
Heat energy given by metal $=$ Heat energy gained by water $+$ Heat energy gained by calorimeter
$468 = m_{w} \times 8.4 + 42$
$m_{w} = \frac{468 - 42}{8.4}$
$= 50.7 g$
Therefore, the mass f water used in the calorimeter is $50.7 g$ .

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