Question

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.

Answer 1

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 × c × change in temperature

= 2100 × 4.2 × (75 - 25)

= 2100 × 4.2 × 50

= 4,41,000

Heat energy is taken by m kg ice to melt into the water at 0°C

= m_i × L

= m_i × 336

Heat energy is taken by water at 0°C to raise it's temperature to 25°C

= m_i × c × change in temperature

= m_i × 4.2 × (25 - 0)

= m_i × 4.2 × 25

= m_i × 105

heat energy released = heat energy taken

Substituting the values we get,

4,41,000 = (m_i × 336) + (m_i × 105)

`⇒` 4,41,000 = 336m_i + 105m_i

`⇒` 4,41,000 = 441m_i

`⇒` m_i = `441000/441`

`⇒` m_i = 1000 g or 1 Kg

Hence, the mass of ice added is 1 Kg.