Question

30 g of ice at 0°C is used to bring down the temperature of a certain mass of water at 70°C to 20°C. Find the mass of water. [Specific heat capacity of water = 4.2 Jg°C⁻¹ and specific latent heat of ice = 336 Jg⁻¹.]

Answer 1

According to the principle of calorimetry, 

Heat lost by water = Total heat gained by ice

`"Mass" xx "Fall in temp"_"water" = "Mass" xx "Latent heat" + "Mass" xx "SHC" xx "Rise in temp"_"cold water"`

m_W × C_W × Δt = m_i × L + m_i × W × Δt

m_W × 4.2 × (70 − 20) = 30 × 336 + 30 × 4.2 × (20 − 0)

m_W × 4.2 × 50 = (30 × 336) + (30 × 4.2 × 20)

m_W × 210 = (10080) + (2520)

m_W × 210 = 12600

Mass of water (m_W) = 60 g

= `60/1000` kg

= 0.06 kg

∴ The mass of water is 0.06 kg.