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Standard VIII

Chemistry

Latent Heat of Vapourisation and Fusion

If 10 gram ...

Question

If $10$ gram of ice at $0^{\circ} C$ is mixed with $10$ gram of water at $40^{\circ} C$ . The final mass of water in mixture is
(Latent heat of fusion of ice $= 80 c a l / g m$ ; specific heat of water $= 1 c a l / g m^{\circ} C$ )

10 gram

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15 gram

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18 gram

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20 gram

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Solution

The correct option is B 15 gram
Heat water can release $= 10 \times 1 \times 40 = 400 c a l$
Heat required by ice to melt $= 800 c a l [80 \times 10]$
$∵$ heat release is less than heat required
So, water left $=$ ice melts $+$ log of water
$= \frac{400}{80} + 10 g = 15 g m$

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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

]

What will be the final temperature, when $150 g m$ of ice at $0^{\circ} C$ is mixed with $300 g m$ of water at $50^{\circ} C$ . Specific heat of water = $1 c a l / g m /^{\circ} C$ . Latent heat of fusion of ice = $80 c a l / g m$

1 kg

of ice at

- 10^{\circ} C

is mixed with

4.4 kg

of water at

30^{\circ} C

. The final temperature of mixture is

Given specific heat of ice

= 0.5 {cal/g}^{\circ} C

,
Specific heat of water

= 1 {cal/g}^{\circ} C

,
Latent heat of fusion for ice

= 80 cal/g

120 g

of ice at

0^{\circ} C

is mixed with

100 g

of water at

80^{\circ} C

. Latent heat of fusion is

80 cal/g

and specific heat of water is

1 {cal/g}^{\circ} C

. The final temperature of the mixture is
(correct answer + 1, wrong answer - 0.25)

Q. What will be the final temperature when

150 g

of ice at

0^{\circ} C

is mixed with

300 g

of water at

50^{\circ} C

? Specific heat of water

= 1 c a l / g^{\circ} C

. Latent heat of fusion of ice

= 80 c a l / g

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If 10 gram of ice at 0∘C is mixed with 10 gram of water at 40∘C. The final mass of water in mixture is (Latent heat of fusion of ice =80cal/gm; specific heat of water =1cal/gm∘C)

The correct option is B 15 gramHeat water can release =10×1×40=400calHeat required by ice to melt =800cal[80×10]∵ heat release is less than heat requiredSo, water left = ice melts + log of water=40080+10g=15gm

If $10$ gram of ice at $0^{\circ} C$ is mixed with $10$ gram of water at $40^{\circ} C$ . The final mass of water in mixture is
(Latent heat of fusion of ice $= 80 c a l / g m$ ; specific heat of water $= 1 c a l / g m^{\circ} C$ )

The correct option is B 15 gram
Heat water can release $= 10 \times 1 \times 40 = 400 c a l$
Heat required by ice to melt $= 800 c a l [80 \times 10]$
$∵$ heat release is less than heat required
So, water left $=$ ice melts $+$ log of water
$= \frac{400}{80} + 10 g = 15 g m$