The sum of the heats gained is zero (some will lose heat).
masswater*c*(Tf-25)+120*Hf+ 120*c*(tf-0)=0
solve for tf
A 500 mL bottle of water, which is at 25 degrees C, is poured over 120 g of ice at -8 degrees C. What will be the final temperature of the water when all the ice has melted. Assume that container is insulated and does not change temperature.
5 answers
What do Tf and Hf stand for? Are they constants? Also, is Tf different from tf?
Tf is final temperature. Hf is the latent heat of fusion for ice. Tf and tf are the same, final temp.
Does c stand for temperature in celsius?
You would have to account for the new water created by the melting ice also.
m*c*(Tf-To)+m*Hf+m*c*(Tf-To)+m*c*(Tf-To)=0
Where the first product is the warming of the ice, which would first increase from -8 to a final T of 0.
The second product is the melting of that same ice that was warmed from -8 to 0.
The third is that new water that was produced by the melted ice (120g), which has an initial temp of 0. Its Tf will be the same as the final temp of all the melted water.
Then the last product is the heat equation from the original water, which has an initial T of 25.
120g*2.11*(0-(-4))+ 120g(333.55)+ 120g(4.18)(Tf-0) + 500g(4.18)(Tf-25)=0
m*c*(Tf-To)+m*Hf+m*c*(Tf-To)+m*c*(Tf-To)=0
Where the first product is the warming of the ice, which would first increase from -8 to a final T of 0.
The second product is the melting of that same ice that was warmed from -8 to 0.
The third is that new water that was produced by the melted ice (120g), which has an initial temp of 0. Its Tf will be the same as the final temp of all the melted water.
Then the last product is the heat equation from the original water, which has an initial T of 25.
120g*2.11*(0-(-4))+ 120g(333.55)+ 120g(4.18)(Tf-0) + 500g(4.18)(Tf-25)=0
3 answers