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3 years ago

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(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, an

d 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.

1 answer:

True [87]3 years ago

7 0

Answer:

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

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