4C.13
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Re: 4C.13
For this problem, we need to solve for the final temperature, which is Tf.
First we write and expression for the q of ice. In this we need to have the enthalpy of fusion as ice is going through a phase change and then a term for the heat that comes from increasing T. This becomes (50.0 g/18.01 g)(6.01*10^3 J.mol-1) + (50 g)(4.184 J/g/C)(Tf-0).
For the water, we just need to have the q=mc*delta T term because it is not going through a phase change. This becomes q=(400 g)(4.184 J/g/C)(Tf-45).
Since we know the final heat of the ice and water is equal to each other, we set both these equations equal and solve for Tf. My final answer was 31 degrees Celsius.
Hope this helps!
First we write and expression for the q of ice. In this we need to have the enthalpy of fusion as ice is going through a phase change and then a term for the heat that comes from increasing T. This becomes (50.0 g/18.01 g)(6.01*10^3 J.mol-1) + (50 g)(4.184 J/g/C)(Tf-0).
For the water, we just need to have the q=mc*delta T term because it is not going through a phase change. This becomes q=(400 g)(4.184 J/g/C)(Tf-45).
Since we know the final heat of the ice and water is equal to each other, we set both these equations equal and solve for Tf. My final answer was 31 degrees Celsius.
Hope this helps!
Re: 4C.13
You can write an expression q1 + q2 = -q3 where q1 is the energy absorbed by the ice to undergo a phase change. q1 can be found through using the enthalpy of fusion value of water and the mass of ice given: 50g / 18.01g * 6.01 kJ mol^-1 = 16685 J
For q2 you can just use the normal q = mCdeltaT equation for how much energy the liquid absorbs after undergoing phase change.
FOr q3, also use q=mCdeltaT
Now plug everything into the first expression and solve for T final
For q2 you can just use the normal q = mCdeltaT equation for how much energy the liquid absorbs after undergoing phase change.
FOr q3, also use q=mCdeltaT
Now plug everything into the first expression and solve for T final
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- Posts: 103
- Joined: Fri Sep 24, 2021 6:13 am
Re: 4C.13
ashna kumar 3k wrote:For this problem, we need to solve for the final temperature, which is Tf.
First we write and expression for the q of ice. In this we need to have the enthalpy of fusion as ice is going through a phase change and then a term for the heat that comes from increasing T. This becomes (50.0 g/18.01 g)(6.01*10^3 J.mol-1) + (50 g)(4.184 J/g/C)(Tf-0).
For the water, we just need to have the q=mc*delta T term because it is not going through a phase change. This becomes q=(400 g)(4.184 J/g/C)(Tf-45).
Since we know the final heat of the ice and water is equal to each other, we set both these equations equal and solve for Tf. My final answer was 31 degrees Celsius.
Hope this helps!
Thank you so much!
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- Posts: 103
- Joined: Fri Sep 24, 2021 6:13 am
Re: 4C.13
kylanjin wrote:You can write an expression q1 + q2 = -q3 where q1 is the energy absorbed by the ice to undergo a phase change. q1 can be found through using the enthalpy of fusion value of water and the mass of ice given: 50g / 18.01g * 6.01 kJ mol^-1 = 16685 J
For q2 you can just use the normal q = mCdeltaT equation for how much energy the liquid absorbs after undergoing phase change.
FOr q3, also use q=mCdeltaT
Now plug everything into the first expression and solve for T final
This really helps! Thank you!
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