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Standard X

Physics

Calorimeter

If 10 g of ic...

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

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Solution

The correct option is B $0$
Given,
$M = 40 g$
$m = 10 g$
$s = 4.2 \times 10^{3} J / k g K$
$L = 3.36 \times 10^{5} J / k g$

Heat loss from $15^{0} C$ to $0^{0} C$
$H_{1} = M S Δ T$
$H_{1} = \frac{40}{1000} \times 4.2 \times 10^{3} \times (15 - 0)$

$H_{1} = 2520 J$

$H_{2} = m L$
$H_{2} = \frac{10}{1000} \times 3.36 \times 10^{5}$
$H_{2} = 3360 J$

From the above calculations,
$H_{2} > H_{1}$
So, the temperature of the mixture is zero degree.
The correct option is D.

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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}

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 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}

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}

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}

]

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If 10 g of ice is added to 40 g 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 )

The correct option is B 0Given,M=40gm=10gs=4.2×103J/kgKL=3.36×105J/kgHeat loss from 150C to 00CH1=MSΔTH1=401000×4.2×103×(15−0)H1=2520JH2=mLH2=101000×3.36×105H2=3360JFrom the above calculations, H2>H1So, the temperature of the mixture is zero degree.The correct option is D.

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