Question

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 Jg^-I°C^-1, Specific latent heat of fusion = 336 Jg^-1]

Answer 1

Given, for water

mass (m_w) = 85 g

Initial temperature = 30°C

Final temperature = 5°C

For Ice, mass= m_i

Initial temperature = 0°C

Final temperature = 5°C

Now

Heat lost by water = m_w × c × Δt

= 8.5 ×  4.2 × (30 - 5)

= 85 ×  4.2 ×  25

= 8925 J

Heat gained by ice = m_i × L + m_i × c × Δt

= m_i × 336 + m_i × 4.2 × (5 - 0)

= 336 m_i + m_i × 4.2 ×  5

= 336 m_i + 21 m_i

= 357 m_i

By principle of calorimetry

Heat lost by water = Heat gained by ice

8925 = 357 m_i

m_i = `8925/357`

= 25 g

Hence, mass of ice required is 25 g.