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 A- 40- CB- 0- CC- 80- CD- -0- C

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

Calorimetry

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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(correct answer + 1, wrong answer - 0.25)

100 g

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540 g of ice at $0^{\circ} C$ is mixed with 540 g of water at $80^{\circ} C$ . The final temperature of the mixture is

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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