A piece of ice of mass 100 g is dropped into 450 g of water at 45°C. Calculate the final temperature of water after all of the ice has melted. (Specific heat capacity of water 4200 Jkg^-1°C^-1, specific heat of fusion of ice = 336 x 10³ Jkg^-1)

Given,

m_i = 100 g

Converting g to kg

1000 g = 1 kg

So, 100 g = $\dfrac{100}{1000}$ = 0.1 kg

m_w = 450 g

Converting g to kg

1000 g = 1 kg

So, 450 g = $\dfrac{450}{1000}$ = 0.45 kg

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 45° C to t° C
= m x c x Δt
= 0.45 × 4200 × (45 – t)
= 85050 – 1890t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 0.1 × 336 x 10³ J
= 33600 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x Δt
= 0.1 × 4200 × (t – 0)
= 0.1 × 4200 × t
= 420t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$33600 + 420t = 85050 - 1890t \\[0.5em] 1890t + 420t = 85050 - 33600 \\[0.5em] 2310t = 51450 \\[0.5em] t = \dfrac{51450}{2310} \\[0.5em] t = 22.27° C$

Hence, final temperature = 22.27°C