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Standard XII

Physics

Specific Heat Capacity

If 10g of i...

Question

If $10 g$ of ice is added to $40 g$ of water at $15^{o} C$ , then the temperature of the mixture is (specific heat of water $= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}$ ,Latent heat of fusion of ice $= 3.36 \times 10^{5} J k g^{- 1}$ )

15^{o} C

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12^{o} C

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10^{o} C

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0^{o} C

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Solution

The correct option is D $0^{o} C$
Heat lost by water to come from $15^{\circ} C t o 0^{\circ} C$ is
$H_{1} = m s △ T$

$\Rightarrow \frac{40}{1000} \times (4.2 \times 10^{3}) \times (15 - 0)$

$\Rightarrow 2520 J$

Now, the heat required to convert $10 g$ ice into $10 g$ water at $0^{o} C$ is :

$H_{2} = m L$

$\Rightarrow \frac{10}{1000} \times (3.36 \times 10^{5})$

$\Rightarrow 3360 J$

Since $H_{2} > H_{1}$ , so the whole ice will not be converted into water, where the temperature of the whole water will be $0^{o} C$

Therefore the temperature of the mixture is $0^{o} C$

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Similar questions

Q. A piece of ice of mass

40 g

0^{o} C

is added to

200 g

of water at

50^{\circ} C

. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

and specific latent heat of fusion of ice =

336 \times 10^{3} J k g^{- 1}

Q. A copper calorimeter of mass

100 g m

contains

200 g m

of a mixture of ice and water. Steam at

100^{\circ} C

under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to

50^{\circ} C

. If the mass of the calorimeter and its contents is now

330 g m

, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper

= 0.42 \times 10^{3} J k g^{- 1} K^{- 1}

,
Specfic heat capacity of water

= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}

Specific heat of fusion of ice

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

Latent heat of condensation of steam

= 22.5 \times 10^{5} J k g^{- 1}

10 g

of ice at

0^{o} C

is mixed with

100 g

of water at

50^{o} C

in a calorimeter. The final temperature of the mixture is [Specific heat of water

= 1 c a l g^{- 1}^{o} C^{- 1}

, latent heat of fusion of ice

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

40 g

of ice at

0^{o} C

is used to bring down the temperature of certain mass of water at

60^{o} C t o 10^{o} C

. Find the mass of water used.
[Specific heat capacity of water

= 4200 J K g^{- 1} C^{- 1}

[Specific latent heat of fusion of ice

= 336 \times 10^{3} J k g^{- 1}

]

Q. Find the result of mixing

10 g

of ice at

- 10^{o} C

with

10 g

of water at

10^{o} C

. Specific heat capacity of ice

= 2.1 J g^{- 1} K^{- 1}

, specific latent heat of ice

= 336 J g^{- 1}

and specific heat capacity of water

= 4.2 J g^{- 1} K^{- 1}

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If 10g of ice is added to 40g of water at 15oC, then the temperature of the mixture is (specific heat of water=4.2×103Jkg−1K−1,Latent heat of fusion of ice =3.36×105Jkg−1)

If $10 g$ of ice is added to $40 g$ of water at $15^{o} C$ , then the temperature of the mixture is (specific heat of water $= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}$ ,Latent heat of fusion of ice $= 3.36 \times 10^{5} J k g^{- 1}$ )