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Expansion of Liquids

A lump of 0.1...

Question

A lump of $0.10 k g$ of ice at $- 10^{\circ} C$ is put in $0.15 k g$ of water at $20^{\circ} C$ . How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water $= 1 k c a l k g^{- 1}$ , specific heat of ice $= 0.50 k c a l k g^{- 1}$ while its latent heat $= 80 k c a l k g^{- 1}$ )

Ice is

68.75 g m

, and water is

181.25 g m

0^{\circ} C

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Ice is

16.75 g m

, and water is

11.25 g m

0^{\circ} C

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Ice is

108.75 g m

, and water is

101.25 g m

10^{\circ} C

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Ice is

6.75 g m

, and water is

181.25 g m

100^{\circ} C

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Solution

The correct option is A Ice is $68.75 g m$ , and water is $181.25 g m$ at $0^{\circ} C$ .
The heat, which $0.15 k g$ of water can release when its temperature is changed from $20^{\circ} C$ to $0^{\circ} C$ .
$Q_{1} = m_{W} s_{W} Δ T_{W}$
where, $m_{W}$ is the mass of the water, $s_{W}$ is the specific heat of water and $Δ T_{W}$ is the change in temperature of water.
Given, $m_{W} = 0.15 k g$ , $s_{W} = 1 k c a l k g^{- 1}$ .
$Q_{1} = 0.15 \times (1 \times 10^{3}) \times (20 - 0)$
$= 3000 c a l$ ... $(1)$
Now, heat absorbed by $0.10 k g$ of ice at $- 10^{\circ} C$ to increase its temperature to $0^{\circ} C$ .

$Q_{2} = m_{i} S_{i} Δ T_{i}$
where $m_{i}$ is the mass of the ice, $s_{i}$ is the specific heat of ice and $Δ T_{i}$ is the change in temperature of ice.
Given, $m_{i} = 0.10 k g$ , $s_{i} = 0.50 k c a l k g^{- 1}$
$Q_{2} = 0.10 \times (0.5 \times 10^{3}) \times [0 - (- 10)]$
$= 500 c a l$
So, remaining heat,
$Q = Q_{1} - Q_{2} = 3000 - 500 = 2500 c a l$

Now as latent heat of ice is $l = 80 k c a l k g^{- 1}$ , the remaining heat will melt ice only
$L = \frac{Q}{m} \Rightarrow m = \frac{Q}{m} = \frac{2500}{80} = 31.25 g$ of ice.
Initial amount of ice $= 0.10 k g = 100 g$
So, the remaining ice,
$= 100 - 31.25 = 68.75 g$
Initial amount of water $= 0.15 k g = 150 g$
Therefore, total water $= 150 + 31.25 = 181.25 g$
The temperature of the system will be $0^{\circ} C$ .

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Similar questions

How much heat energy is released when 5.0 g of water at $20^{o} C$ changes into ice at $0^{o} C$ ? Take specific heat capacity of water = 4.2 J $g^{-} 1 K^{- 1}$ , specific latent heat of fusion of ice = 336 J $g^{- 1}$ .

Q. A lump of

0.10 k g

of ice at

- 10^{\circ} C

is put in

0.15 k g

of water at

20^{\circ} C

. How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water

= 1 k c a l k g^{- 1}

, specific heat of ice

= 0.50 k c a l k g^{- 1}

while its latent heat

= 80 k c a l k g^{- 1}

)

Q. A lump of

0.10 k g

of ice at

- 10^{\circ} C

is put in

0.15 k g

of water at

20^{\circ} C

. How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water

= 1 k c a l k g^{- 1}

, specific heat of ice

= 0.50 k c a l k g^{- 1}

while its latent heat

= 80 k c a l k g^{- 1}

)

A lump of 0.10 kg of ice at -10°C is put in 0.15 kg of water at 20°C.How much water and ice will be found in the mixture when it has reached thermal equilibrium? Specific heat capacity of ice=0.50 kcal/kg and its latent heat of melting=80 kcal/kg

Q. A lump of ice (0.1 kg) at

- 10^{\circ}

C is put in 0.15 kg of water at

20^{\circ}

C. How much water will be found in the mixture when it has reached thermal equilibrium?

(C_{i c e} = 0.5 k c a l k g^{- 1} k^{- 1}, L_{i c e} = 80 k c a l k g^{- 1})

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A lump of 0.10kg of ice at −10∘C is put in 0.15kg of water at 20∘C. How much water and ice will be found in the mixture when it has reached in thermal equilibrium. (Specific heat of water =1kcalkg−1, specific heat of ice =0.50kcalkg−1 while its latent heat =80kcalkg−1)