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.
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
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
To find the final temperature of the water when all the ice has melted, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the ice to reach its melting point. The specific heat capacity of ice is 2.09 J/g·°C.
Q1 = mass * specific heat capacity * change in temperature
Q1 = 120 g * 2.09 J/g·°C * (0 °C - (-8 °C))
Q1 = 120 g * 2.09 J/g·°C * 8 °C
Q1 = 2006.4 J
So, the ice will absorb 2006.4 Joules of heat energy to melt completely.
Next, we need to calculate the heat lost by the water to reach the final temperature. The specific heat capacity of water is 4.18 J/g·°C.
Q2 = mass * specific heat capacity * change in temperature
Q2 = 500 mL * 1 g/mL * 4.18 J/g·°C * (Tf °C - 25 °C)
Q2 = 2090 J * (Tf °C - 25 °C)
Q2 = 2090 Tf - 52250 J
Since no heat is lost or gained by the container, the heat gained by the ice (Q1) equals the heat lost by the water (Q2).
Q1 = Q2
2006.4 J = 2090 Tf - 52250 J
Let's solve for Tf, the final temperature of the water:
2006.4 J + 52250 J = 2090 Tf
54256.4 J = 2090 Tf
Tf = 54256.4 J / 2090
Tf ≈ 26 °C
Therefore, the final temperature of the water when all the ice has melted will be approximately 26 degrees Celsius.