Answer:
Total mass of ice = 38.06g
Explanation:
Since the heat capacity of calorimeter is negligible.
The water is already at 0°C, so the heat loss can no longer reduce the temperature of the water. It is used for fusion and forming more ice.
The equilibrium temperature will be 0°C, because the heat gain by ice is only enough to bring it down to 0°C.
Heat gained by ice = heat loss by water
Heat gained by ice (from -14°C to 0°C) = heat lost to fusion by water (heat of fusion of some amount of the water present in the calorimeter)
mi Ci ∆Ti = mw . L ......1
Where;
mi = mass of ice = 35g = 0.035 kg
Ci = specific heat capacity of ice = 2090 J/kg ∙ K
∆Ti = change in temperature of ice = 0-(-14) = 14 K
mw = the mass of water that have gained enough heat for fusion ( mass of water converted to ice)
L = latent heat of fusion of water = 33.5 × 10^4 J/kg.
From equation 1;
mw = (mi Ci ∆Ti )/L
mw = (0.035×2090×14)/335000
mw = 0.00306 kg
mw = 3.06 g
Therefore, 3.06 g of water has been converted to ice.
When combined with the initial amount of ice initially in the calorimeter (at 0°C)
Total mass of ice = mi + 3.06g = 35g + 3.06g = 38.06g
Total mass of ice = 38.06g
