Question

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is 2090 J/kg ∙ K, that of water is 4186 J/kg ∙ K, and the latent heat of fusion of water is 33.5 × 10^4 J/kg.

Answer 1

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

Answer 2

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