Question

600-g of ice at −15℃ is in a calorimeter when 100 g of water at 20℃ is added to it. Water cools down and part of it freezes as a result. Find the mass of the remaining liquid water. Neglect the heat capacity of the calorimeter. [$c_i$ = 0.5 cal/g/℃ = 2090 J/kg/℃, $c_w$ = 1 cal/g/℃ = 4186J/kg/℃, $L_f$ = 79.7 cal/g = 3.33 × 10⁵ J/kg]

Answer 1

Answer:

$\Delta m =68.654\ g$

Explanation:

Given:

• mass of ice, $m_i=600\ g$

• initial temperature of ice, $T_{ii}=-15\ ^{\circ}C$

• mass of water added, $m_w=100\ g$

• initial temperature of water, $T_{iw}=20\ ^{\circ}C$

• specific heat capacity of ice, $c_i=2.090\ J.g^{-1}.^{\circ}C^{-1}$

• specific heat capacity of water, $c_w=4.186\ J.g^{-1}.^{\circ}C^{-1}$

• latent heat of fusion, $L_f=333\ J.g^{-1}$

According to question a part of water freezes while the other remains liquid:

So,

final temperature of the mixture, $T_f=0 ^{\circ}C$

Now using the equation of heat:

Heat required by the total ice of given temperature to melt completely:

$Q_i=m_i.c_i.\Delta T$

$Q_i=600\times 2.090\times (0-(-15))$

$Q_i=18810\ J$ .....................................(1)

Heat released by the total mass of given water to come to freezing point:

$Q_w=m_w.c_w.\Delta T$

$Q_w=100\times 4.186\times (20)$

$Q_w=8372\ J$ ...............................................(2)

Now from the eq. (1) & (2) the difference in heat is equivalent to the mass of water frozen.

$\Delta Q=Q_i-Q_w$

$\Delta Q=18810-8372$

$\Delta Q=10438\ J$

Now the mass of water frozen:

$\Delta Q = m.L_f$

$m=\frac{\Delta Q}{L_f}$

$m=\frac{10438}{333}$

$m=31.345\ g$ of more ice is formed from the water.

Mass of remaining liquid:

$\Delta m =m_w-m$

$\Delta m =100-31.345$

$\Delta m =68.654\ g$