What is the final temperature, in oC, after a 27.0 g piece of ice, at 0oC, is placed in a styrofoam cup with 120 g of water initially at 70.5oC.
Assume there is no transfer of heat to or from the surroundings.
The specific heat of water = 4.184 JK-1g-1
The heat of fusion of ice = 333 Jg-1
gain of heat on melting ice + gain of heat on raising T from 0C to Tfinal + heat lost by 120 g H2O at 70.5C to Tfinal = 0
Each [ ] gives the three processes above.
[mass ice x heat fusin] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific heat H2O x (Tfinal-Tinitial) = 0
Substitute the numbers and solve for the only unknown of Tfinal.
To find the final temperature, we need to calculate the heat lost by the water and gained by the ice.
First, let's calculate the heat lost by the water using the formula:
Q = mcΔT
Where:
Q is the amount of heat lost or gained (in Joules),
m is the mass of the water (in grams),
c is the specific heat of water (in J/g°C),
ΔT is the change in temperature (in °C).
Given:
m = 120 g (mass of water)
c = 4.184 J/g°C (specific heat of water)
ΔT = 70.5°C (initial temperature of water - final temperature)
Qwater = mcΔT
Qwater = (120 g) * (4.184 J/g°C) * (70.5°C - ΔT)
Next, let's calculate the heat gained by the ice using the formula:
Q = mHf
Where:
Q is the amount of heat lost or gained (in Joules),
m is the mass of the ice (in grams),
Hf is the heat of fusion of ice (in J/g).
Given:
m = 27.0 g (mass of ice)
Hf = 333 J/g (heat of fusion of ice)
Qice = mHf
Qice = (27.0 g) * (333 J/g)
Since there is no transfer of heat to or from the surroundings, the heat lost by the water is equal to the heat gained by the ice:
Qwater = Qice
Substituting the respective formulas:
(120 g) * (4.184 J/g°C) * (70.5°C - ΔT) = (27.0 g) * (333 J/g)
Simplifying the equation:
(120 * 4.184 * (70.5 - ΔT)) = (27.0 * 333)
Now, we can solve for ΔT:
(50208 - 501.888ΔT) = 8991
-501.888ΔT = 8991 - 50208
-501.888ΔT = -41217
ΔT = -41217 / -501.888
ΔT ≈ 82.09°C
Finally, to find the final temperature, we subtract ΔT from the initial temperature of the water:
Final temperature = (70.5°C) - (82.09°C)
Final temperature ≈ -11.59°C
Therefore, the final temperature, in oC, after the ice is placed in the styrofoam cup with water is approximately -11.59°C.