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Standard VII

Physics

Conduction

Mixing is bei...

Question

Mixing is being done of $m$ $g m$ of steam at $100^{0} C$ is mixed with $31.50 g m$ ice at $- 20^{0} C .$ Take $s_{i c e} = s_{s t e a m} = 0.5 c a l / g m^{0} C,$ $s_{w a t e r} = 1 c a l / g m^{0} C$ and the latent heat of fusion and vaporization is $80 c a l / g m$ and $540 c a l / g m$ respectively.The final temperature is found to be $10^{0} C$ . Find the value of $m$ .

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Solution

Heat lost by steam is equal to Heat gained by ice
Heat absorbed by ice $=$ Heat used to convert ice from $- 20^{0} C$ to $0^{0} C$ $+$ Heat to convert ice at $0^{0} C$ to water at $0^{0} C$ + Heat to convert water at $0^{0} C$ to water at $10^{0} C$
Heat released by steam $=$ Heat used to convert steam at $100^{0} C$ to water at $100^{0} C$ $+$ Water at $100^{0} C$ to water at $10^{0} C$ .

$31.5 \times 0.5 \times 20 + 31.5 \times 80 + 31.5 \times 1 \times 10 = m \times 540 + m \times 1 \times 90$
$∴ 630 \times m = 31.5 \times 100$
$∴ m = 5 g m$

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Similar questions

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

]

20 gm

ice at

- 10^{\circ} C

is mixed with

m gm

steam at

100^{\circ} C

. The minimum value of

m

so that finally all ice and steam converts into water is

[

Use, specific heat and latent heat as

C_{i c e} = 0.5 cal/gm^{\circ} C, C_{water} = 1 cal/gm^{\circ} C, L_{m e l t}

= 80 cal/gm

and

L_{v a p o r}

= 540 cal/gm

]

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

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

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

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]

Q. 20 gm ice at

- 10^{o} C

is mixed with m gm steam at

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. Minimum value of m so that finally all ice and steam converts into water. (Use

S_{i c e} = 0.5

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S_{w a t e r} = 1

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, L(melting)=80 cal/gm and L (vapourization)=540 cal/gm)

Q. In an insulated vessel,

0.05 k g

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and

0.45 k g

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253 K

are mixed. Then, find the final temperature of the mixture.
Given,

L_{f u s i o n} = 80 c a l / g = 336 J / g

L_{v a p o r i s a t i o n} = 540 c a l / g = 2268 J / g

S_{i c e} = 2100 J / k g K = 0.5 c a l / g K

and

S_{w a t e r} = 4200 J / k g K = 1 c a l / g K

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Mixing is being done of m gm of steam at 1000C is mixed with 31.50gm ice at −200C.Take sice=ssteam=0.5cal/gm0C, swater=1cal/gm0Cand the latent heat of fusion and vaporization is 80cal/gm and 540cal/gm respectively.The final temperature is found to be 100C. Find the value of m.