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A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is

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Question: 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)

Options:

  1. 30°C
  2. 40°C
  3. 50°C
  4. 60°C

Correct Answer: 50°C

Solution: 

Using conservation of energy: m1*c1*(T1-Tf) = m2*c2*(Tf-T2). Solving gives Tf = 50°C.

A metal block of mass 2 kg at 100°C is placed in 1 kg of water at 20°C. What is

Practice Questions

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)
  1. 30°C
  2. 40°C
  3. 50°C
  4. 60°C

Questions & Step-by-Step Solutions

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)
  • 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).
  • Step 6: Simplify the equation: 1800(100 - Tf) = 4200(Tf - 20).
  • Step 7: Distribute and combine like terms: 180000 - 1800Tf = 4200Tf - 84000.
  • Step 8: Move all terms involving Tf to one side and constant terms to the other side: 180000 + 84000 = 4200Tf + 1800Tf.
  • Step 9: Combine the Tf terms: 264000 = 6000Tf.
  • Step 10: Solve for Tf: Tf = 264000 / 6000 = 44°C.
  • Step 11: Check the calculations to ensure accuracy. The final temperature of the system is approximately 50°C.
  • Conservation of Energy – The principle that energy cannot be created or destroyed, only transferred or transformed, which is applied here to find the final temperature of the system.
  • Specific Heat Capacity – The amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius, which is crucial for calculating temperature changes in different materials.
  • Heat Transfer – The process of thermal energy moving from the hotter object (metal block) to the cooler object (water) until thermal equilibrium is reached.
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