Solved 2024 Question Paper ICSE Class 10 Physics — Question 16

Given,

For Water

Mass of water (m_w) = 85 g

Initial temperature (T_i) = 30°C

Final temperature (T_f) = 5°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) = 5°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

= 85 × 4.2 × (30 - 5)

= 85 × 4.2 × 25 = 8925 J

Heat gained by ice = m_iL_i + m_ic_w△t

= m_i × 336 + m_i × 4.2 × (5-0)

= 336m_i + m_i × 4.2 × 5

= 336m_i + 21m_i = 357m_i

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 8925 = 357 m_i

⟹ m_i = $\dfrac{8925}{357}$ = 25 g

So, the mass of required ice is 25 g.