Consider 60 g of hot water at 70 degree C poured into a cavity in a very large block of ice at 0 degree C. A)What will be the final temperature of the water in the cavity?
B)Find what mass of ice must melt in order to cool the hot water down to this temperature
To find the final temperature of the water in the cavity, we can use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the ice that melts.
Let's break this down into two parts:
A) Final temperature of the water in the cavity:
1. First, let's find the heat lost by the hot water. We can use the formula Q = mcΔT, where Q is the heat, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Q = (60 g) * (4.18 J/g·°C) * (70°C - Tf), where Tf is the final temperature of the water.
Note: The specific heat capacity of water is 4.18 J/g·°C.
2. Next, let's find the heat gained by the ice that melts. We can use the formula Q = mL, where Q is the heat, m is the mass of the ice that melts, and L is the latent heat of fusion of ice.
Q = (m g) * (333.55 J/g), where mg is the mass of the ice that melts.
Note: The latent heat of fusion of ice is 333.55 J/g.
3. Since the heat lost by the hot water is equal to the heat gained by the ice, we can set up the equation:
(60 g) * (4.18 J/g·°C) * (70°C - Tf) = (m g) * (333.55 J/g)
4. Simplify the equation and solve for Tf:
250.8 J/g·°C * (70°C - Tf) = 333.55 J/g * mg
(70°C - Tf) = (333.55 J/g * mg) / 250.8 J/g·°C
Tf = 70°C - (333.55 J/g * mg) / 250.8 J/g·°C
B) Mass of ice that must melt:
To find the mass of ice that must melt in order to cool the hot water down to the final temperature, we need to substitute the final temperature (Tf) we found above into the initial equation and solve for mg.
Now, let's substitute the value of Tf into the initial equation:
(60 g) * (4.18 J/g·°C) * (70°C - Tf) = (m g) * (333.55 J/g)
Substitute Tf = 70°C - (333.55 J/g * mg) / 250.8 J/g·°C:
(60 g) * (4.18 J/g·°C) * (70°C - (333.55 J/g * mg) / 250.8 J/g·°C) = (m g) * (333.55 J/g)
Simplify the equation and solve for mg:
(60 g) * (4.18 J/g·°C) * (70°C * 250.8 J/g·°C) = (333.55 J/g * mg)
99720 J = 333.55 J/g * mg
mg = 99720 J / (333.55 J/g)
mg = 299 g
Therefore,
A) The final temperature of the water in the cavity would be approximately 0.97°C.
B) To cool the hot water down to this temperature, approximately 299 grams of ice must melt.