A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, Specific heat of metal = 0.9 J/g°C)
Practice Questions
1 question
Q1
A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, Specific heat of metal = 0.9 J/g°C)
30°C
40°C
50°C
60°C
Using conservation of energy: m1*c1*(T1-Tf) = m2*c2*(Tf-T2). Solving gives Tf = 50°C.
Questions & Step-by-step Solutions
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Q
Q: A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, Specific heat of metal = 0.9 J/g°C)
Solution: Using conservation of energy: m1*c1*(T1-Tf) = m2*c2*(Tf-T2). Solving gives Tf = 50°C.
Steps: 11
Step 1: Identify the masses and specific heats of the metal block and water. The mass of the metal block (m1) is 2 kg, and its specific heat (c1) is 0.9 J/g°C. The mass of the water (m2) is 1 kg, and its specific heat (c2) is 4.2 J/g°C.
Step 2: Convert the masses from kg to grams because the specific heats are given in J/g°C. So, m1 = 2000 g and m2 = 1000 g.
Step 3: Write down the initial temperatures. The initial temperature of the metal block (T1) is 100°C, and the initial temperature of the water (T2) is 20°C.
Step 4: Set up the equation using the principle of conservation of energy. The heat lost by the metal block equals the heat gained by the water: m1*c1*(T1-Tf) = m2*c2*(Tf-T2).
Step 5: Substitute the values into the equation: 2000 g * 0.9 J/g°C * (100°C - Tf) = 1000 g * 4.2 J/g°C * (Tf - 20°C).
