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
