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Standard XII

Physics

Specific Heat Capacity

A calorimeter...

Question

A calorimeter contains ice. Find the water equivalent of the calorimeter if $500 cal$ are required to heat it together with its contents from $270 K$ to $272 K$ and $16600 cal$ from $272$ to $274 K$ . Take specific heat of water as $1 c a l / g m^{o} C$ and that of ice as $0.5 c a l / g m^{o} C$ .

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Solution

We have

$500 c a l = m S Δ T$

$500 c a l = m \times 0.5 \times 2$

$m = 500 g$

Similarly

$16600 c a l = m \times 0.5 \times 1 + m L + m \times 1 \times 1$

$16600 = m (1.5 + L)$

$16600 = 500 (1.5 + L)$

$L = 31.7 c a l / g$

Thus latent heat of fusion is $31.7 c a l / g$ .

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Similar questions

5 g m

of water at

30^{\circ} C

and

5 g m

of ice at

- 20^{\circ} C

are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice

= 0.5 c a l / g m^{\circ} C

and latent heat of ice

= 80 c a l / g m

50 g m

ice at -

10^{0} C

is mixed with

20 g m

steam at

100^{0} C .

When the mixture finally reaches its steady state inside a calorimeter of water equivalent

1.5 g m

then : [Assume calorimeter was initially at

0^{0} C,

Take latent heat of vaporization of water

= 540

cal/gm, Latent heat of fusion of water

= 80

cal/gm, specific heat capacity of water

= 1 c a l / g m -^{0} C,

specific heat capacity of ice

= {0.5 cal/ gm}^{0} C

]

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}

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 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

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A calorimeter contains ice. Find the water equivalent of the calorimeter if 500cal are required to heat it together with its contents from 270K to 272K and 16600 cal from 272 to 274K . Take specific heat of water as 1cal/gmoC and that of ice as 0.5cal/gmoC.

We have 500cal=mSΔT 500cal=m×0.5×2 m=500g Similarly 16600cal=m×0.5×1+mL+m×1×1 16600=m(1.5+L) 16600=500(1.5+L) L=31.7cal/g Thus latent heat of fusion is 31.7cal/g.

A calorimeter contains ice. Find the water equivalent of the calorimeter if $500 cal$ are required to heat it together with its contents from $270 K$ to $272 K$ and $16600 cal$ from $272$ to $274 K$ . Take specific heat of water as $1 c a l / g m^{o} C$ and that of ice as $0.5 c a l / g m^{o} C$ .

We have

$500 c a l = m S Δ T$

$500 c a l = m \times 0.5 \times 2$

$m = 500 g$

Similarly

$16600 c a l = m \times 0.5 \times 1 + m L + m \times 1 \times 1$

$16600 = m (1.5 + L)$

$16600 = 500 (1.5 + L)$

$L = 31.7 c a l / g$

Thus latent heat of fusion is $31.7 c a l / g$ .