A 1 kg block of metal at 100°C is placed in 2 kg 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) (2021)
Practice Questions
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Q1
A 1 kg block of metal at 100°C is placed in 2 kg 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) (2021)
25°C
30°C
35°C
40°C
Using conservation of energy: m1*c1*(T_initial - T_final) = m2*c2*(T_final - T_initial). Solving gives T_final = 35°C.
Questions & Step-by-step Solutions
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Q
Q: A 1 kg block of metal at 100°C is placed in 2 kg 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) (2021)
Solution: Using conservation of energy: m1*c1*(T_initial - T_final) = m2*c2*(T_final - T_initial). Solving gives T_final = 35°C.
Steps: 10
Step 1: Identify the masses and specific heats of the materials involved. The mass of the metal block (m1) is 1 kg, and its specific heat (c1) is 0.9 J/g°C. The mass of the water (m2) is 2 kg, and its specific heat (c2) is 4.18 J/g°C.
Step 2: Convert the masses from kg to grams because the specific heats are given in J/g°C. So, m1 = 1000 g (1 kg) and m2 = 2000 g (2 kg).
Step 3: 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 4: Set up the equation using 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 5: Substitute the known values into the equation: 1000 g * 0.9 J/g°C * (100°C - T_final) = 2000 g * 4.18 J/g°C * (T_final - 20°C).
