100 g of ice latent heat 80 cal g-1- at 0-C is mixed with 100 g of water specific heat 1 cal g-1-C at 80-C- The final temperature of the mixture will be

We need nbsp;m L_f=100 × 80=8000 cal to convert ice at 0^0 C to water at nbsp;0^0 C But for water at 80^0 C to go to ice at 0^0 C it releases nbsp; ...

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Byju's Answer

Standard XII

Physics

Specific Heat Capacity

100 g of ice ...

Question

$100 g$ of ice (latent heat $80 {cal g}^{- 1}$ ), at $0^{\circ} C$ is mixed with $100 g$ of water (specific heat $1 {cal g}^{- 1} /^{\circ} C$ ) at $80^{\circ} C$ . The final temperature of the mixture will be

0^{\circ} C

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40^{\circ} C

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80^{\circ} C

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< 0^{\circ} C

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Solution

The correct option is A $0^{\circ} C$
We need $m L_{f} = 100 \times 80 = 8000 c a l$ to convert ice at $0^{0} C$ to water at $0^{0} C$

But for water at $80^{0} C$ to go to ice at $0^{0} C$ it releases
$\begin{matrix} m S_{w} Δ T & = 100 \times 1 \times (80 - 0) = 8000 c a l \end{matrix}$

SInce both are same, mixture will be at Zero degrees.

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Similar questions

100 g

of ice

(latent heat 80 {cal g}^{- 1})

0^{\circ} C

is mixed with

100 g

of water (specific heat

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

) at

80^{\circ} C

. The final temperature of the mixture will be:

100 g

of ice (latent heat

80 {cal g}^{- 1}

), at

0^{\circ} C

is mixed with

100 g

of water (specific heat

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

) at

80^{\circ} C

. The final temperature of the mixture will be

100 g

of ice (latent heat

80 c a l g^{- 1}

,at

0^{o} C

) is mixed with

100 g

of water (specific heat 1 cal

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

) at

80^{o} C

. The final temperature of the mixture will be:

Q. Work done in converting one gram of ice at

- 10^{\circ} C

into steam at

100^{\circ} C

is -
[Given, specific heat of ice

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

, specific heat of water

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

, latent heat of fusion of ice

L_{f} = 80 cal/g

, & latent heat of vaporization of steam

L_{v} = 540 cal/g

]

Q. Steam at

100 ° C

is passed into

20 g

of water at

10 ° C

. When water acquires a temperature of

80 ° C

, the mass of water present will be: [Take specific heat of water

= 1 {cal g}^{- 1} C^{- 1}

and latent heat of steam

= 540 {cal g}^{- 1}

]

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100 g of ice (latent heat 80 cal g−1), at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1/∘C ) at 80∘C. The final temperature of the mixture will be

The correct option is A 0∘CWe need mLf=100×80=8000 cal to convert ice at 00 C to water at 00 C But for water at 800 C to go to ice at 00 C it releases mSwΔT=100×1×(80−0)=8000 cal SInce both are same, mixture will be at Zero degrees.

$100 g$ of ice (latent heat $80 {cal g}^{- 1}$ ), at $0^{\circ} C$ is mixed with $100 g$ of water (specific heat $1 {cal g}^{- 1} /^{\circ} C$ ) at $80^{\circ} C$ . The final temperature of the mixture will be