Question

How much ice (at 0°C) must be added to 1.90 kg of water at 79 °C so as to end up with all liquid at 8 °C? (ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf= 3.35 Ă— 10 5 J/kg, Lv= 2.26 Ă— 10 6 J/kg) kg

Answer 1

m = mass of the ice added = ?

M = mass of water = 1.90 kg

$c_{w}$ = specific heat of the water = 4186 J/(kg ⁰C)

$c_{i}$  = specific heat of the ice = 2000 J/(kg ⁰C)

$L_{f}$ = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

$T_{ii}$  = initial temperature of ice = 0 ⁰C

$T_{wi}$ = initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m $c_{w}$  (T - $T_{ii}$ ) + m $L_{f}$  = M $c_{w}$  ($T_{wi}$ - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg