Can someone please give me a hand on this problem? Two 20.0-g ice cubes at -11.0
ID: 863223 • Letter: C
Question
Can someone please give me a hand on this problem?
Two 20.0-g ice cubes at -11.0 ?C are placed into 265 g of water at 25.0 ?C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts. Number ? C heat capacity of H20(s) 37.7 J/(mol- K) heat capacity of H20(0 75.3 J/(mol? K) enthalpy of fusion of H2O 6.01 kJ/mol One way to solve this type of problem is to express the total heat exchange as one giant formula. In this case, the final temperature is the only unknown value. heat needed to warm 40.0 g of ice to 0 ?C heat needed to heat needed to warm = heat released to cool melt 40.0 g of ice 40.0 g of water to the 265 g of water to the final temperature final temperatureExplanation / Answer
The heat gained by the ice cubes and heat is lost by water.
Q ice = - Q water.
Let us calculate Q ice.
Step 1: Heat gained when ice at -11 oC is converted to 0 oC ice
Q = m x c x delta T
m = moles of ice
mass of ice = 2 x 20 = 40 g
Moles of ice = 40 / 18 = 2.22 mol
Tf = 0 oC = 273 K
Ti = -11 oC = 262 K
Q = 2.22 g x 37.7 x (273 - 262)
Q = 921 J
Step 2: Heat gained by 0 oC ice to 0 oC water
Q = m x heat of fusion
Q = 2.22 x 6.01
Q = 13.3422 kJ = 13342.2 J
Step 3: Heat gained when 0 oC water to final temperature.
Q = m c delta T
Q = 2.22 x 75.3 x [Tf - 273]
Q = 167.2 [Tf - 273]
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Q water
265 g of water at 25 oC is changed to final temperature.
moles = 265 / 18 = 14.72 mol
Q = m c delta T
Q = 14.72 x 75.3 x [Tf - 298]
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Q ice = -Q water
921 + 13342.2 + 167.2 [Tf - 273] = - [14.72 x 75.3 x [Tf - 298]]
167.2Tf - 31382.4 = -1108.42Tf + 330307.96
1275.62Tf = 361690.36
Tf = 287.59 K
T(in Kelvin) = T (in degree C) + 273
T (in degree C) = T(in Kelvin) - 273
T (in oC) = 287.59 - 273
T = 14.5 oC