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What mass of ice at 0°C added to 2.1 kg water, will cool it down from 75°C to 25°C?

Given: Specific heat capacity of water = 4.2 Jg-1 C-1

Specific latent heat of ice = 336 Jg-1

Calorimetry

ICSE Sp 2024

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Answer

Given,

mw = 2.1 kg = 2100 g

Specific heat capacity of water = 4.2 Jg-1 C-1

Specific latent heat of ice = 336 Jg-1

mi = ?

Heat energy given out by water in lowering it's temperature from 75° C to 25° C
= m x c x change in temperature
= 2100 x 4.2 x (75 - 25)
= 2100 x 4.2 x 50
= 4,41,000

Heat energy taken by m kg ice to melt into water at 0° C
= mi x L
= mi x 336

Heat energy taken by water at 0° C to raise it's temperature to 25° C
= mi x c x change in temperature
= mi x 4.2 x (25 - 0)
= mi x 4.2 x 25
= mi x 105

heat energy released = heat energy taken

Substituting the values we get,

4,41,000 = (mi x 336) + (mi x 105)

⇒ 4,41,000 = 336mi + 105mi

⇒ 4,41,000 = 441mi

⇒ mi = 441000441\dfrac{441000}{441}

⇒ mi = 1000 g or 1 Kg

Hence, the mass of ice added = 1 Kg

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