You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Molar and Specific Heat Capacity

Calculate the...

Question

Calculate the amount of ice, which is sufficient to cool $150$ of water, contained in a vessel of mass $50$ g(S.H.C. $0.4$ $J g^{- 1}^{o} C^{- 1}$ at $30^{o}$ C , such that the final temperature of mixture is $5^{o} C$ .

Open in App

Solution

Let the required amount of ice be $m$
Heat gain = Heat loss
$m \times 80 + m \times 1 \times 5 = 150 \times 1 (30 - 5) + 50 \times 0.4 \times (30 - 5) \Rightarrow 85 m = 4250 \Rightarrow m = 50 g$

Suggest Corrections

Similar questions

Q. Calculate the amount of ice, which is sufficient to cool 150 g of water, contained in a vessel of mass 50 g (S.H.C.

0.4 J g^{- 1}^{o} C^{- 1}

) at

30^{o} C

, such that the final temperature of the mixture is

5^{o} C

Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30°C, such that the final temperature of the mixture is 5^oC. Take specific heat capacity of material of vessel as 0.4 J g^-1 ^oC^-1 ; specific latent heat of fusion of ice = 336 J g^-1 ; specific heat capacity of water = 4.2 J g^-1 °C^-1.

Q. Calculate the mass of ice needed to cool

150

g of water contained in a calorimeter of mass

50

g at

32^{o}

C such that the final temperature is

5^{o}

C.
Specific heat capacity of calorimeter

= 0.4 J / g^{o} C

Specific heat capacity of water

= 4.2 J / g^{o} C

Latent heat capacity of ice

= 330 J / g

Q. Ice at

- 20^{\circ} C

is added to

50 g

of water at

40^{\circ} C

, when the temperature of the mixture reaches

0^{\circ} C

, it is found that

20 g

of ice is still unmelted. The amount of ice added to the water was close to

(Specific heat of water = 4.2 {Jg}^{- 1}^{\circ} C^{- 1} Specific heat of Ice = 2.1 {Jg}^{- 1}^{\circ} C^{- 1} Heat of fusion of water at 0^{\circ} C = 334 {Jg}^{- 1})

In a laboratory experiment for finding specific latent heat of ice, 100 g of water at 30^oC was taken in a calorimeter made of copper and of mass 10 g. When 10 g of ice at 0^oC was added to the mixture and kept within the liquid till the ice melted completely, the final temperature of the mixture was found to be 20^oC.

Calculate the value of the latent heat of fusion of ice from the data discussed above.

Join BYJU'S Learning Program

Calculate the amount of ice, which is sufficient to cool 150 of water, contained in a vessel of mass 50g(S.H.C. 0.4 Jg−1oC−1 at 30oC , such that the final temperature of mixture is 5oC.

Let the required amount of ice be mHeat gain = Heat lossm×80+m×1×5=150×1(30−5)+50×0.4×(30−5)⇒85m=4250⇒m=50g

Calculate the amount of ice, which is sufficient to cool $150$ of water, contained in a vessel of mass $50$ g(S.H.C. $0.4$ $J g^{- 1}^{o} C^{- 1}$ at $30^{o}$ C , such that the final temperature of mixture is $5^{o} C$ .

Let the required amount of ice be $m$
Heat gain = Heat loss
$m \times 80 + m \times 1 \times 5 = 150 \times 1 (30 - 5) + 50 \times 0.4 \times (30 - 5) \Rightarrow 85 m = 4250 \Rightarrow m = 50 g$