Question

Question: A kilogram of water at 20°C is placed in a refrigerator. How much heat be extracted before all the water turns to ice?
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Answer 1

Answer :-

$\sf\huge\boxed{100000\: calories}$

Solution :-

We know that,

Latent heat of ice = 80 Cal g-¹

Mass of water (m) => 1 kg = 1000 g

Latent heat = 80 cal g-¹

T¹ = 20° C ; T² = 0° C

Therefore,

Rise in temperature => (20 - 0)°C = 20°C

We know that,

Heat extracted (Q) = Mass (m) × Specific heat capacity (s) × Rise/Fall in temperature (t)

=> 1000 × 20 = 20,000 Calories

Now,

Heat extracted from water to freeze into ice = m × L

= 1000 × 80

= 80,000 cal

Therefore, Total heat extracted :-

= 20,000 + 80,000

= 100000 cal

Answer 2

Thanks for asking the question!

$\huge\bf{Answer:-}$

Refer the above attachment.

Hope it helps!

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