A piece of ice of mass 40 g is added to 200 g of water at 50° C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of fusion of ice = 336 x 10³ J kg^-1.

Given,

m_i = 40 g

m_w = 200 g

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

specific latent heat of fusion of ice = 336 x 10³ J kg^-1

final temperature of water = ?

Let final temperature = t

Heat energy given out by water when it cools from 50° C to t° C
= m x c x change in temperature
= 200 × 4.2 × (50 – t)
= 42000 – 840t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 40 × 336 J
= 13440 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x change in temperature
= 40 × 4.2 × (t – 0)
= 40 × 4.2 × t
= 168t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$13440 + 168t = 42000 – 840t \\[0.5em] 168t + 840t = 42000 – 13440 \\[0.5em] 1008t = 28560 \\[0.5em] t = \dfrac{28560}{1008} \\[0.5em] t = 28.330° C$

Hence, final temperature = 28.330° C