A calorimeter of mass $50 g$ specific heat capacity $0.42 J g^{- 1} {}^{o}{C^{- 1}}$ 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. (Take specific heat capacity of the metal piece $0.3 J g^{- 1} {}^{o}{C^{- 1}}$ , specific heat capacity of water $4.2 J g^{- 1} {}^{o}{C^{- 1}}$ )

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Solution

Step 1: Given data
Mass of calorimeter $m_{1} = 50 g$
Since the substance of the calorimeter and the water is kept in equilibrium initially, the temperature of the water will be equal to that of the calorimeter. Therefore,
The initial temperature of the calorimeter $T_{1} = 20^{o} C$ and
The initial temperature of the water $T_{1} = 20^{o} C$
Specific heat capacity of the calorimeter $s_{1} = 0.42 J g^{- 1} {}^{o}{C^{- 1}}$
Mass of metal piece $m_{2} = 20 g$
Initial temperature of the metal piece $T_{2} = 100^{o} C$
Specific heat capacity of metal piece $s_{2} = 0.3 J g^{- 1} {}^{o}{C^{- 1}}$
Specific heat capacity of water $s_{3} = 4.2 J g^{- 1} {}^{o}{C^{- 1}}$
The final equilibrium temperature of all three substances is $T = 22^{o} C$ .
Step 2: Write expressions for heat changes in the system:
Heat lost by the metal piece = $m_{2} s_{2} (T_{2} - T)$
Heat gained by calorimeter = $m_{1} s_{1} (T - T_{1})$
Heat gained by water = $m_{3} s_{3} (T - T_{1})$
Step 3: Apply the principle of calorimetry-
We know that the principle of calorimetry says that the heat lost by the hotter object is equal to the heat gained by the colder object when two objects at different temperatures are mixed together.
In this situation, the metal piece is at high temperature than the water and the calorimeter. Therefore,
Heat lost by metal pipe = Heat gained by calorimeter + heat gained by the water
Step 4: Solve the equation for the mass of water using the given data-
By substituting values, we get
$20 \times 0.3 (100 - 22) = 50 \times 0.42 (22 - 20) + 4.2 m_{3} (22 - 20)$
or, $426 = 8.4 m_{3}$
or, $m_{3} = 50.71 g$
Hence, the mass of water used in the calorimeter is $50.71 g$ .

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Similar questions

Q. 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

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}

]

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

(Take, specific heat capacity of metal piece = $0.3 J . g^{- 1} . {}^{\circ}C^{- 1}$ and specific heat capacity of water = $0.42 J . g^{- 1} . {}^{\circ}C^{- 1}$ )

Calculate the mass of ice needed to cool $150 g$ of water contained in a calorimeter of mass $50 g$ at $32^{o} C$ such that the final temperature is $5^{o} C$ .
Specific heat capacity of calorimeter $0.4 J g^{- 1} {}^{o}{C^{- 1}}$
Specific heat capacity of water $4.2 J g^{- 1} {}^{o}{C^{- 1}}$
Latent heat capacity of ice $330 J g^{- 1}$