The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of ice = 336000 J kg^-1.

Given,

m_w = 170 g = 0.17 kg

specific heat capacity of water = 4200 J kg^-1 K^-1

specific latent heat of ice = 336000 J kg^-1

m_i = ?

Heat energy given out by water in lowering it's temperature from 50° C to 5° C
= m x c x change in temperature
= 0.17 x 4200 x (50 - 5)
= 0.17 x 4200 x 45
= 32,130

Heat energy taken by m kg ice to melt into water at 0° C
= m_i x L
= m_i x 336000

Heat energy taken by water at 0° C to raise it's temperature to 5° C
= m_i x c x change in temperature
= m_i x 4200 x (5 - 0)
= m_i x 4200 x 5
= m_i x 21000

heat energy released = heat energy taken

Substituting the values we get,

$32130 = (m$i \times 336000 ) + (mi \times 21000 ) \\[0.5em] 32130 = mi \times 357000 \\[0.5em] \Rightarrow mi = \dfrac{32130}{357000} \\[0.5em] \Rightarrow m_i = 0.09 \text{ Kg} = 90 \text{ g}

Hence, the mass of ice added = 90 g