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Physics

85 g of water at 30°C is cooled to 5°C by adding certain mass of ice. Find the mass of ice required.

[Specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹ , Specific latent heat of fusion = 336 J g⁻¹]

Calorimetry

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Answer

Given,

For Water

Mass of water (mw) = 85 g

Initial temperature (Ti) = 30°C

Final temperature (Tf) = 5°C

Specific heat capacity of water (cw) = 4.2 J g⁻¹ °C⁻¹

For ice

Initial temperature (Ti) = 0°C

Final temperature (Tf) = 5°C

Specific latent heat of fusion (Li) = 336 J g⁻¹

Let,

Mass of ice = (mi)

Now,

Heat lost by water = mwcw△t

= 85 × 4.2 × (30 - 5)

= 85 × 4.2 × 25 = 8925 J

Heat gained by ice = miLi + micw△t

= mi × 336 + mi × 4.2 × (5-0)

= 336mi + mi × 4.2 × 5

= 336mi + 21mi = 357mi

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 8925 = 357 mi

⟹ mi = 8925357\dfrac{8925}{357} = 25 g

So, the mass of required ice is 25 g.

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