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

Given,

m_w = 2.1 kg = 2100 g

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

Specific latent heat of ice = 336 Jg^-1

m_i = ?

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
= m_i x L
= m_i x 336

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

heat energy released = heat energy taken

Substituting the values we get,

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

⇒ 4,41,000 = 336m_i + 105m_i

⇒ 4,41,000 = 441m_i

⇒ m_i = $\frac{441000}{441}$

⇒ m_i = 1000 g or 1 Kg

Hence, the mass of ice added = 1 Kg