Respuesta :
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
Answer:
403 g
Explanation:
mass of block of ice = 35 g
initial ice temperature = -14⁰c
mass of water = 400 g
water temperature = 0⁰c
How much ice is left when the system reaches equilibrium
specific heat of ice = 2090 j/kg
specific heat of water = 4186 j/kg
Latent heat of fusion of water = 33.5 * 10^4 J/kg
heat required to melt the ice to equilibrium ( Q ) from -14 to 0 degrees Celsius
Q(ice) = mass of ice * specific heat capacity of water * temperature
= 35 * 2090 * 14 = 1024100 j
Q (water ) = Mw * Lf
= 400 * 33.5 * 10^4 = 134000000 j
Total energy required to meet equilibrium = 135024100 J
LETS ASSUME TOTAL MASS = 400 + 35 after reaching equilibrium at 0 degrees
to calculate mass left in the Calorimeter
M*Lf = 135024100
there mass left in Calorimeter = 135024100/ Lf ( 33.5 10^4)
MASS LEFT = 403 g
