One gram of ice at 0 - C is added to 5 yarn of water at 10 - C - If the latent heat of ice be 80 cal -g- then the final temperature of the mixture is

The correct option is B 0 ^ ∘ C Heat required when 1 g ice becomes 1 g water =mL =1 × 80 calHeat released when 5 kg water at ...

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Byju's Answer

Standard VI

Physics

Thermal Equilibrium

One gram of i...

Question

One gram of ice at $0^{\circ} C$ is added to 5 yarn of water at $10^{\circ} C$ . If the latent heat of ice be 80 cal /g, then the final temperature of the mixture is

5^{\circ} C

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0^{\circ} C

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- 5^{\circ} C

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None of the above

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Solution

The correct option is B $0^{\circ} C$
Heat required when $1 g$ ice becomes $1 g$ water $= m L$
$= 1 \times 80$ cal
Heat released when $5 k g$ water at $10^{0} C$ turns ice $= m c o s θ$
$= 5 \times 1 \times 10$
$= 50 J$
Since more heat is required for ice to turn water
Hence final mixture would be ice temperature $0^{0} C$

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One gram of ice at 0∘C is added to 5 yarn of water at 10∘C . If the latent heat of ice be 80 cal /g, then the final temperature of the mixture is

The correct option is B 0∘CHeat required when 1g ice becomes 1g water =mL=1×80calHeat released when 5kg water at 100C turns ice =mcosθ=5×1×10=50JSince more heat is required for ice to turn waterHence final mixture would be ice temperature 00C

One gram of ice at $0^{\circ} C$ is added to 5 yarn of water at $10^{\circ} C$ . If the latent heat of ice be 80 cal /g, then the final temperature of the mixture is