A 100 g piece of metal at 100°C is placed in 200 g of water at 20°C. What will b
Practice Questions
Q1
A 100 g piece of metal at 100°C is placed in 200 g of water at 20°C. What will be the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.5 J/g°C) (2023)
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Questions & Step-by-Step Solutions
A 100 g piece of metal at 100°C is placed in 200 g of water at 20°C. What will be the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.5 J/g°C) (2023)
Step 1: Identify the mass and specific heat of the metal and water. The mass of the metal is 100 g and its specific heat is 0.5 J/g°C. The mass of the water is 200 g and its specific heat is 4.2 J/g°C.
Step 2: Set the initial temperatures. The initial temperature of the metal is 100°C and the initial temperature of the water is 20°C.
Step 3: Use the heat transfer principle. The heat lost by the metal will equal the heat gained by the water. This can be written as: (mass of metal) * (specific heat of metal) * (change in temperature of metal) = (mass of water) * (specific heat of water) * (change in temperature of water).
Step 4: Define the change in temperature. Let the final temperature be T. The change in temperature for the metal is (100 - T) and for the water is (T - 20).
Step 5: Substitute the values into the equation: 100 g * 0.5 J/g°C * (100 - T) = 200 g * 4.2 J/g°C * (T - 20).
Step 9: Solve for T: 21800 = 890T, so T = 21800 / 890.
Step 10: Calculate T to find the final temperature: T = 24.5°C.
Heat Transfer – The principle of conservation of energy where heat lost by the metal equals heat gained by the water.
Specific Heat Capacity – The amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
Thermal Equilibrium – The state reached when two substances at different temperatures come into contact and exchange heat until they reach the same temperature.
