(a) Two 44 g ice cubes are dropped into 220 g of water in a thermally insulated container. If the water is initially at 24°C, and the ice comes directly from a freezer at -19°C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? The specific heat of water is 4186 J/kg·K. The specific heat of ice is 2220 J/kg·K. The latent heat of fusion is 333 kJ/kg.

A.) 16.45 degree

B.) 19.86 degree

Explanation:

The parameters given in the question are:

Mass of the ice = 2 × 44g = 88g

Mass of water = 220g

Water temperature T1 = 24 degrees

Initial ice temperature TC = - 19 degrees

The specific heat of water is 4186 J/kg·K. The specific heat of ice is 2220 J/kg·K. The latent heat of fusion is 333 kJ/kg.

When the ice were dropped into the water, the ice gain heat and the water loses heat to the ice.

From the conservative of energy

Heat gain by the ice = heat lost by the water.

That is,

ML + MC Øi = MC Øw

Where Ø = change in temperature

88 × 333 + 88 × 2220 ( T + 19 ) = 220 × 4186 ( 24 - T )

29304 + 195360(T + 19) = 920920(24 -T)

Open the bracket

29304 + 195360T + 3711840 = 22102080 - 920920T

3741144 + 195360T = 22102080 - 920920T

Collect the like terms

195360T + 920920T = 22102080 - 3741144

1116280T = 18360936

T = 18360936 / 1116280

T = 16.45 degree

(b) What is the final temperature if only one ice cube is used?

Using the same formula

ML + MCØ = MCØ

44 × 333 + 44 × 2220 ( T + 19 ) = 220 × 4186 ( 24 - T )

14652 + 97680(T + 19) = 920920(24 -T)

Open the bracket

14652 + 97680T + 1855920 = 22102080 - 920920T

1870572 + 97680T = 22102080 - 920920T

Collect the like terms

97680T + 920920T = 22102080 - 1870572

1018600T = 20231508

T = 20231508 / 1018600

T = 19.86 degree

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

lol

explanation: