A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. What will be the final temperature of the system assuming no heat loss? (2022)
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 will be the final temperature of the system assuming no heat loss? (2022)
30°C
40°C
50°C
60°C
Using the principle of conservation of energy: m1*c1*(T_initial1 - T_final) = m2*c2*(T_final - T_initial2). Solving gives T_final = 50°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 will be the final temperature of the system assuming no heat loss? (2022)
Solution: Using the principle of conservation of energy: m1*c1*(T_initial1 - T_final) = m2*c2*(T_final - T_initial2). Solving gives T_final = 50°C.
Steps: 6
Step 1: Identify the masses and initial temperatures of the two objects. The metal block has a mass (m1) of 1 kg and an initial temperature (T_initial1) of 100°C. The water has a mass (m2) of 2 kg and an initial temperature (T_initial2) of 20°C.
Step 2: Write down the specific heat capacities. For simplicity, we can use c1 for the metal and c2 for water. The specific heat capacity of water is typically 4.18 J/g°C, but we will keep it as c2 for now.
Step 3: Use the conservation of energy principle. This means that the heat lost by the metal block will equal the heat gained by the water. We can express this as: m1 * c1 * (T_initial1 - T_final) = m2 * c2 * (T_final - T_initial2).
Step 4: Substitute the known values into the equation. We have: 1 kg * c1 * (100°C - T_final) = 2 kg * c2 * (T_final - 20°C).
Step 5: Rearrange the equation to solve for T_final. This involves isolating T_final on one side of the equation.
Step 6: Solve the equation. After performing the calculations, we find that T_final = 50°C.
