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 gm ,and water is 181.25 gm at

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Solution

The correct option is A Ice is 68.75 gm ,and water is 181.25 gm at .
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

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

10gm of ice at $- 20^{\circ}$ is dropped into a calorimeter containing 10gm of water at $10^{\circ} C$ , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain.