$10 g$ of ice at $- 20^{\circ} C$ is added to $20 g$ of water at $20^{\circ} C$ . Assuming that no heat is lost to the surrounding, find the temperature of the mixture.
[Take, specific heat capacity of ice $C_{i} = 2100 {J kg}^{- 1}^{\circ} C^{- 1}$ , specific heat capacity of water $C_{w} = 4200 {J kg}^{- 1}^{\circ} C^{- 1}$ & Latent heat of fusion of ice $L = 3 .36 \times 10^{5} {J kg}^{- 1}$ ]

0^{\circ} C

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1^{\circ} C

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2^{\circ} C

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3^{\circ} C

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Solution

The correct option is A $0^{\circ} C$
The heat released by water when it cools down $20^{\circ} C$ to $0^{\circ} C$ is given by
$Q_{1} = m C_{w} Δ T$
$Q_{1} = 0.02 kg \times 4200 {J kg}^{- 1}^{\circ} C^{- 1} \times 20^{\circ} C = 1680 J . . . . (1)$
The heat absorbed by ice to reach $0^{\circ} C$ from $- 20^{\circ} C$ is given by
$Q_{2} = m^{'} C_{i} Δ T$
$Q_{2} = 0.01 kg \times 2100 {J kg}^{- 1}^{\circ} C^{- 1} \times 20^{\circ} C = 420 J . . . . . (2)$
Also:
Heat required to just melt $10 g$ of ice is given by
$Q_{3} = m^{'} L = 0.01 kg \times 3.36 \times 10^{5} {J kg}^{- 1} = 3360 J . . . . . (3)$

From the principle of calorimetry,
Heat lost by water = Heat gained by ice.
$Q_{1} = Q_{2} + Q_{3}$
But from $(1)$ , $(2)$ and $(3)$ , we can say that,
$Q_{1} < Q_{2} + Q_{3}$
[ $∵ Q_{1} < Q_{3}$ and $Q_{1} > Q_{2}$ ]
Thus, we can conclude that all the ice will not be melt and temperature becomes $0^{\circ} C$
Hence, option (a) is the correct answer.

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Similar questions

10 g

of ice at

- 20^{\circ} C

is added to

20 g

of water at

20^{\circ} C

. Assuming that no heat is lost to the surrounding, find the final temperature of the mixture.
[Take, specific heat capacity of ice

C_{i} = 2100 {J kg}^{- 1}^{\circ} C^{- 1}

, specific heat capacity of water

C_{w} = 4200 {J kg}^{- 1}^{\circ} C^{- 1}

& Latent heat of fusion of ice

L = 3 .36 \times 10^{5} {J kg}^{- 1}

]

Q. A 5g piece of ice at

- 20^{\circ}

C is put into 10g of water at

30^{\circ}

C. Assuming that heat is enchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice

= 2100 J k g^{- 1}^{\circ} C^{- 1}

. Specific heat capacity of water

= 4200 J k g^{- 1}^{\circ} C^{- 1}

and latent heat of fusion of ice

3.36 \times 10^{5} J k g^{- 1}

A 5 g piece of ice at $- 20^{\circ}$ C is put into 10 g of water at $30^{\circ}$ C. Assuming that heat is exchanged only between the ice and the water. Then find the heat required for phase change. Specific heat capacity of ice $= 2100 J k g^{- 1}^{\circ} C^{- 1}$ . Specific heat capacity of water $= 4200 J k g^{- 1}^{\circ} C^{- 1}$ and latent heat of fusion of ice $3.36 \times 10^{5} J k g^{- 1}$ .

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

Q. The amount of heat energy required to convert