What will be the final temperature- when 150 gm of ice at 0- C is mixed with 300 gm of water at 50- C- Specific heat of water -1 cal - gm - C- Latent heat of fusion of ice -80 cal - gmA- 4-8- CB- -3- CC- 0- CD- 6-7- C

The correct option is D 6.7°C Let us assume that T 0^∘C Heat lost by water = Heat gained by ice to melt + heat by water formed from ice ⇒ 300 × 1 × (50 T ...

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Byju's Answer

Standard VI

Physics

Thermal Equilibrium

What will be ...

Question

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 = $80 c a l / g m$

- 3°C

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0°C

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4.8°C

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6.7°C

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Solution

The correct option is D
6.7°C

Let us assume that $T > 0^{\circ} C$

Heat lost by water = Heat gained by ice to melt + heat by water formed from ice

$\Rightarrow$ $300 \times 1 \times (50 - T)$ = $150 \times 80 + 150 \times 1 \times (T - 0)$

$\Rightarrow$ $T$ = ${6.7}^{\circ} C$

Hence our assumption that $T$ $>$ $0^{\circ} C$ is correct.

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Similar questions

Q. What will be the final temperature when

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(correct answer + 1, wrong answer - 0.25)

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What will be the final temperature, when 150 gm of ice at 0∘Cis mixed with 300 gm of water at 50∘C. Specific heat of water = 1cal/gm/∘C. Latent heat of fusion of ice = 80 cal/gm

The correct option is D 6.7°C Let us assume that T > 0∘C Heat lost by water = Heat gained by ice to melt + heat by water formed from ice ⇒ 300 × 1 × (50−T) = 150 × 80 + 150 × 1 × (T−0) ⇒ T = 6.7∘C Hence our assumption that T > 0∘C is correct.

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 = $80 c a l / g m$