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 $330 J g^{- 1}$

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Solution

Step 1: Given Data
Mass of water $m_{1} = 150 g$
Initial temperature of water $T_{1} = 32^{o} C$
Specific heat capacity of water $s_{1} = 4.2 J g^{- 1} {}^{o}{C^{- 1}}$
Mass of calorimeter $m_{2} = 50 g$
The initial temperature of the calorimeter $T_{1} = 32^{o} C$
Specific heat capacity of water $s_{2} = 0.4 J g^{- 1} {}^{o}{C^{- 1}}$
The final equilibrium temperature given as $T = 5^{o} C$
Step 2:Calculate the heat lost
We want to add ice in the water so that ice immediately melts and the temperature should become $T = 5^{o} C$ . For this, we will calculate the heat energy required to be released in order to make its temperature from $T_{1} = 32^{o} C$ to $T = 5^{o} C$ .
$heat lost = heat lost by water + heat lost by calorimeter$
Or, $heat lost = m_{1} s_{1} (T_{1} - T) + m_{2} s_{2} (T_{1} - T) = 150 \times 4.2 (32 - 5) + 50 \times 0.4 \times (32 - 5)$
Or, $heat lost = m_{1} s_{1} (T_{1} - T) + m_{2} s_{2} (T_{1} - T) = 150 \times 4.2 (32 - 5) + 50 \times 0.4 \times (32 - 5)$
Or, $heat lost = 17550 J$
Step 3: Calculate the mass of ice
Now this heat lost will be gained by ice. Upon gaining this much heat, ice will immediately melt into water.
Therefore, $mass of required m = \frac{Q}{L}$ , where $L$ is the Latent heat capacity of ice $330 J g^{- 1}$ . And we obtained $heat Q = 17550 J$ .
So, the Amount of heat gain for melting is
$Q_{1} = m L Q_{1} = m \times 330 Q_{1} = 330 m$
The amount of heat gain for changing the temperature of ice is,
$Q_{2} = m s_{1} (T - 0^{\circ}) Q_{2} = m \times 4.2 \times (5^{\circ} - 0^{\circ}) Q_{2} = 21 m$
As the amount of heat gained is equal to the amount of heat lost thus,
$Q_{1} + Q_{2} = h e a t l o s t 330 m + 21 m = 17550 351 m = 17550 m = 50 k g$
Thus, the mass of ice needed to cool the calorimeter in the given situation will be $50 g$ .

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Similar questions

Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at $32^{ο} C$ such that the final temperature is $5^{ο} C$ .

(Take specific heat capacity of water $c_{w} = 4.2 J / g^{ο} C$ and latent heat capacity of ice $L = 330 J / g$ )

Q. 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^{o} C

Specific heat capacity of water

= 4.2 J / g^{o} C

Latent heat capacity of ice

= 330 J / g

Q. A copper vessel of mass 100g contains 150g of water at

50^{o} C

. How much ice is needed to cool it to

5^{o} C

?
Given: Specific heat capacity of copper

= 0.4 J g^{- 1}

^{o} C^{- 1}

Specific heat capacity of water

= 4.2 J g^{- 1}

^{o} C^{- 1}

Specific latent heat of fusion of ice

= 336 J g^{- 1}

Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30°C, such that the final temperature of the mixture is 5^oC. Take specific heat capacity of material of vessel as 0.4 J g^-1 ^oC^-1 ; specific latent heat of fusion of ice = 336 J g^-1 ; specific heat capacity of water = 4.2 J g^-1 °C^-1.

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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Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at 32oC such that the final temperature is 5oC.Specific heat capacity of calorimeter 0.4Jg-1C-1oSpecific heat capacity of water 4.2Jg-1C-1oLatent heat capacity of ice 330Jg-1