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Interconversion of State

Water of mass...

Question

Water of mass $m_{2} = 1 k g$ is contained in a copper calorimeter of mass $m_{1} = 1 k g$ . Their common temperature is $t = 10^{o} C$ . Now a piece of ice of mass $m_{3} = 2 k g$ and temperature $- 11^{o} C$ is dropped into the calorimeter. Neglecting any heat loss, the final temperature of the system is [specific heat of copper $0.1 K c a l / k g^{o} C$ , specific heat of water $= 1 K c a l / k g^{o} C$ , specific heat of ice $= 0.5 K c a l / k g^{o} C$ , latent heat of fusion of ice $= 78.7 K c a l / k g$ ]

0^{o} C

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4^{o} C

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- 4^{o} C

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- 2^{o} C

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Solution

The correct option is B $0^{o} C$
Heat is lost by calorimeter plus water while it is gained by ice.
Heat lost = heat gained
$Q_{l o s t} = Q_{c a l o r i m e t e r} + Q_{w a t e r}$
$= m_{c a l o r i m e t e r} \times s_{c a l o r i m e t e r} \times (10 - 0) + m_{w a t e r} \times s_{w a t e r} \times 1$
$= 1 \times 1 \times 10 + 1 \times 0.1 \times 10$
$= 11 k c a l$
$Q_{g a i n e d} = m_{i c e} \times s_{i c e} \times (0 - (- 11)) = 2 \times 0.5 \times 11 = 11 k c a l$
Thus, all the heat lost by calorie meter and water is used for melting the ice from $- 11^{0} C$ to $0^{0} C$ .
Hence, the final temp of the system will be $0^{0} C$ , ice and water will co-exist at $0^{0} C$ .

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Similar questions

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

Q. Calculate the quantity of heat required to convert

10 k g

of ice at

0^{o} C

to water at

50^{o} C

(Specific heat of water is

4200 J / k g^{o} C

, Latent heat of fusion of ice is

336 \times 10^{3} J / k g

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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 metallic calorimeter of mass 100g contains 200 g of ice and the initial temperature of the calorimeter and the ice is

- 10^{\circ} C

. Heat is supplied to the system containing calorimeter and ice at a constant rate of

50 c a l s^{- 1}

. Find the time required to raise the temperature of the system to

50^{\circ} C

. [Neglect the loss of heat to the surroundings].

Take specific heat capacity of calorimeter and ice as

0.2 c a l g^{- 1}^{\circ} C^{- 1}

and

\frac{1}{2} c a l g^{- 1}^{\circ} C^{- 1}

.Take latent heat of fusion of ice as

80 c a l g^{- 1}

and specific heat capacity of water as

1 c a l g^{- 1}^{\circ} C^{- 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}]

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