A 200 g piece of metal at 100°C is placed in 300 g of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2022)
Practice Questions
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Q1
A 200 g piece of metal at 100°C is placed in 300 g of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2022)
22.5°C
30°C
35°C
40°C
Using the principle of conservation of energy: m1c1(T_initial - T_final) = m2c2(T_final - T_initial). Solving gives T_final = 35°C.
Questions & Step-by-step Solutions
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Q
Q: A 200 g piece of metal at 100°C is placed in 300 g of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.18 J/g°C, specific heat of metal = 0.9 J/g°C) (2022)
Solution: Using the principle of conservation of energy: m1c1(T_initial - T_final) = m2c2(T_final - T_initial). Solving gives T_final = 35°C.
Steps: 9
Step 1: Identify the masses and specific heats of the substances involved. The mass of the metal (m1) is 200 g, its specific heat (c1) is 0.9 J/g°C. The mass of the water (m2) is 300 g, and its specific heat (c2) is 4.18 J/g°C.
Step 2: Write down the initial temperatures. The initial temperature of the metal (T_initial_metal) is 100°C, and the initial temperature of the water (T_initial_water) is 20°C.
Step 3: Set up the equation based on the principle of conservation of energy. The heat lost by the metal will equal the heat gained by the water: m1 * c1 * (T_initial_metal - T_final) = m2 * c2 * (T_final - T_initial_water).
Step 4: Substitute the known values into the equation: 200 g * 0.9 J/g°C * (100°C - T_final) = 300 g * 4.18 J/g°C * (T_final - 20°C).
