If 12.0 gram of ice are dropped in 150. ml of water at 45 degrees celcius, what is the final temperature of the water?
Note the correct spelling of celsius.
I assume the temperature of the ice is zero C.
heat gained by ice + heat gained by melted ice + heat lost by 45 degree water = 0.
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) = 0
Solve for Tfinal. I get something like 36 degrees but that's an approximation.
Filling in the equation, here is what I have. Still not coming up with 36 degrees. Please help:
[mass ice x heat fusion]
12.0g x 334j/g
[mass melted ice x specific heat water x (Tfinal-Tinitial)]
12.0g x 4.18j/g x 45 (Delta T is 45 because temp of ice from 0 degrees to water at 45 degrees)
[mass water x specific heat water x (Tfinal-Tinitial)
150g x 4.18j/g x (Tf - 45)
Also, how would I go about solving for Tfinal?
After some sleep, i was able to come up with the following. It's still not 36 degrees, but close. Can some check my calculation?
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) =0
12.0g x 334j/g = 4008j + 12.0g x 4.18j/gC x 45C =2257j + 150g x 4.18j/gC x (Tf – 45C) = 0
-6265j = 150g x 4.18j/gC x (Tf – 45C)
-6265 = 627C x (Tf – 45C)
-9.99 C = Tf – 45C
Tf = 35.01C
The middle term is the one that is not right. 12.0 g x 4.184 x (Tfinal-Tinitial). Tfinal is Tfinal no matter where you see it; therefore, Tfinal is what you solve for. You have one Tfinal in the middle term and another in the last term. As far as delta T goes for the middle terms, that is not 45 (45-0) but Tfinal-0). The melted ice starts at zero C and goes to Tfinal.
Makes perfect sense. Thank you.
To find the final temperature of the water, we can use the principle of conservation of energy. The heat gained by the water should be equal to the heat lost by the ice as they reach thermal equilibrium.
First, let's calculate the amount of heat gained by the water using the formula:
Q = mcΔT
Where:
Q = heat gained by the water (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature minus initial temperature)
Given:
m = 150.0 g
c = 4.18 J/g°C
Initial temperature of water = 45°C (which can be converted to Celsius by directly using it in the formula since the specific heat capacity of water is in J/g°C)
Qwater = mcΔT
= (150.0 g)(4.18 J/g°C)(Tf - 45°C)
Next, let's calculate the amount of heat lost by the ice. To melt the ice, we need to apply the heat of fusion since the ice is at its melting point.
Qice = m × heat of fusion
Given:
m = 12.0 g
Heat of fusion for ice = 334 J/g
Qice = (12.0 g)(334 J/g)
According to the principle of conservation of energy, Qwater = Qice.
Therefore, we can equate the two equations:
(150.0 g)(4.18 J/g°C)(Tf - 45°C) = (12.0 g)(334 J/g)
Now we can solve for Tf:
(150.0 g)(4.18 J/g°C)(Tf - 45°C) = (12.0 g)(334 J/g)
Simplifying the equation:
625.5(Tf - 45) = 4008
Expanding the equation:
625.5Tf - 28147.5 = 4008
Moving the terms:
625.5Tf = 32155.5
Isolating Tf:
Tf = 32155.5 / 625.5
Calculating Tf by dividing:
Tf ≈ 51.48°C
Therefore, the final temperature of the water is approximately 51.48°C.