Solved 2025 Specimen Paper ICSE Class 10 Physics — Question 16

Given,

For Water

Mass of water (m_w) = 2.1 kg = 2100 g

Initial temperature (T_i) = 75°C

Final temperature (T_f) = 25°C

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

For ice

Initial temperature (T_i) = 0°C

Final temperature (T_f) = 25°C

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

Let,

Mass of ice = (m_i)

Now,

Heat lost by water = m_wc_w△t

= 2100 × 4.2 × (75 - 25)

= 2100 × 4.2 × 50 = 441000 J = 441 kJ

Heat gained by ice = m_iL_i + m_ic_w△t

= m_i × 336 + m_i × 4.2 × (25-0)

= 336m_i + m_i × 4.2 × 25

= 336m_i + 105m_i = 441m_i

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 441000 = 441 m_i

⟹ m_i = $\dfrac{441000}{441}$ = 1000 g = 1 kg

So, the mass of required ice is 1 kg.