A vessel of mass 80 g-S-H-C- - 0-8 J-g- deg-c- contains 250g of water at 35 deg-c-Calculate the amount of ice at 0 deg-c- which must be added to it-so that the final temp- is 5 deg-c- -Sp- latent heat of ice-340J-g-

Dear Student Qgained #160;by #160;ice= #160;Qlost #160;by #160;water+ #160;Q #160;lost #160;by #160;vesselmL=mwatercwater #8710;T+mvesselcv ...

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

Byju's Answer

Standard IX

Physics

Anomalous Expansion of Water

A vessel of m...

Question

A vessel of mass 80 g(S.H.C. = 0.8 J/g/°c) contains 250g of water at 35°c.Calculate the amount of ice at 0°c, which must be added to it,so that the final temp. is 5°c? (Sp. latent heat of ice=340J/g)

Open in App

Solution

Dear Student

$Q_{g a i n e d b y i c e} = Q_{l o s t b y w a t e r} + Q_{l o s t b y v e s s e l} m L = m_{w a t e r} c_{w a t e r} ∆ T + m_{v e s s e l} c_{v e s s e l} ∆ T m (340 J / g) = (250) (4.18) (35 - 5) + (80) (0.8) (35 - 5) 340 m = 31350 + 1920 = 33270 m = 97.85 g$
Regards

Suggest Corrections

Similar questions

Q. 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}

30^{o}

C , such that the final temperature of mixture is

5^{o} C

Q. A vessel of negligible heat capacity contains 40g of ice at 0

^{o}

C. 8g of steam at 100

^{o}

C is passed into the ice to melt it. Find the final temperature of mixture (Latent heat of ice

= 336 J g^{- 1}

, Latent heat of steam

= 2268 J g^{- 1}

. Sp. heat capacity of water

= 4.2 J g^{- 10} C^{- 1}

)

Q. 2 kg of ice at

- 20^{o} C

is mixed with 5 kg of water at

20^{o} C

in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel. It is given that the specific heats of water and ice are

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

and

0.5 k c a l k g^{- 1}^{o} C^{- 1}

respectively and the latent heat of fusion of ice is 80 kcal/kg.

2 kg

of ice at

- 20^{\circ} C

is mixed with

5 kg

of water at

20^{\circ} C

in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are

1 kcal {kg}^{- 1} {^{\circ} C}^{- 1}

and

0.5 kcal {kg}^{- 1} {^{\circ} C}^{- 1}

while the latent heat of fusion of ice is

80 kcal/kg

Q. A calorimeter contains

400 g

of water at a temperature of

5^{\circ} C

. Then,

200 g

of water at a temperature of

+ 10^{\circ} C

and

400 g

of ice at a temperature of

- 60^{\circ} C

are added. What is the final temperature(in

^{o} C

) of the contents of the calorimeter? Specific heat capacity of water

= 1000 c a l / k g / K

. Specific latent heat of fusion of ice

= 80 \times 1000 c a l / k g

. Relative specific heat of ice

= 0.5

Join BYJU'S Learning Program

A vessel of mass 80 g(S.H.C. = 0.8 J/g/°c) contains 250g of water at 35°c.Calculate the amount of ice at 0°c, which must be added to it,so that the final temp. is 5°c? (Sp. latent heat of ice=340J/g)

Dear Student Qgained by ice= Qlost by water+ Q lost by vesselmL=mwatercwater∆T+mvesselcvessel∆Tm(340 J/g)=(250)(4.18)(35-5)+(80)(0.8)(35-5)340 m=31350+1920=33270m =97.85 g Regards

Dear Student

$Q_{g a i n e d b y i c e} = Q_{l o s t b y w a t e r} + Q_{l o s t b y v e s s e l} m L = m_{w a t e r} c_{w a t e r} ∆ T + m_{v e s s e l} c_{v e s s e l} ∆ T m (340 J / g) = (250) (4.18) (35 - 5) + (80) (0.8) (35 - 5) 340 m = 31350 + 1920 = 33270 m = 97.85 g$
Regards