A 63.8 kg block of silver is at initially at 107 C. 1.0 kg of ice at its freezing point is placed onto the block to cool it. As the ice melts, the water is free to flow off (so that the water never increases in temperature). After all the ice melts, what is the final temperature of the silver? Round answers to the nearest whole number, and in degrees Celsius.
Not unless you give us the heat of fusion of water and the specific heat of water and silver.
mass silver * specific heat of silver * (107-T) = heat of fusion of water * mass of ice
To find the final temperature of the silver block, we can use the principle of conservation of energy. The energy lost by the silver block as it cools down will be equal to the energy gained by the ice as it melts. Let's break down the steps to determine the final temperature:
Step 1: Heat gained by ice to melt it
The heat gained by the ice can be calculated using the formula: Q = m * Lf, where Q is the heat gained, m is the mass of ice, and Lf is the latent heat of fusion for water.
Given:
Mass of ice (m) = 1.0 kg
Latent heat of fusion for water (Lf) = 334,000 J/kg
Q = m * Lf
Q = 1.0 kg * 334,000 J/kg
Q = 334,000 J
Step 2: Heat lost by the silver block to cool down
The heat lost by the silver block can be calculated using the formula: Q = m * c * ΔT, where Q is the heat lost, m is the mass of the silver block, c is the specific heat capacity of silver, and ΔT is the change in temperature.
Given:
Mass of silver block (m) = 63.8 kg
Specific heat capacity of silver (c) = 235 J/kg°C
Initial temperature of silver (Ti) = 107°C
Final temperature of silver (Tf) = ?
Q = m * c * ΔT
Q = 63.8 kg * 235 J/kg°C * (107°C - Tf)
Step 3: Conservation of energy
According to the principle of conservation of energy, the heat gained by the ice is equal to the heat lost by the silver block.
334,000 J = 63.8 kg * 235 J/kg°C * (107°C - Tf)
Step 4: Solving for the final temperature
Now we can solve the equation for the final temperature of the silver block (Tf).
334,000 J = 63.8 kg * 235 J/kg°C * (107°C - Tf)
Divide both sides of the equation by (63.8 kg * 235 J/kg°C):
334,000 J / (63.8 kg * 235 J/kg°C) = 107°C - Tf
Simplify:
0.2345 ≈ 107°C - Tf
Subtract 107°C from both sides of the equation:
-106.7655 ≈ -Tf
Multiply both sides by -1:
106.7655 ≈ Tf
Round the final answer to the nearest whole number:
Tf ≈ 107°C
Therefore, the final temperature of the silver block after all the ice melts is approximately 107°C.