Question:

A block of metal with a mass of 2 kg is initially at a temperature of 100°C. The specific heat capacity of the metal is 0.4 J/g·°C. The block is then submerged in a container filled with 5 kg of water at an initial temperature of 20°C. After some time, the final equilibrium temperature of the metal and water is found to be 35°C. Calculate the heat gained by the water and the heat lost by the metal during this process. Assume no heat is lost to the surroundings.

Answer:

To solve this problem, we need to use the equation for heat transfer:

Q = mcΔT

where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the heat gained by the water:

Given:

• Mass of water (m₁) = 5 kg
• Initial temperature of water (T₁) = 20°C
• Final temperature of water (T₂) = 35°C
• Specific heat capacity of water (c₁) = 4.184 J/g·°C (specific heat capacity of water is approximately 4.184 J/g·°C)

Using the formula Q = mcΔT, we can calculate the heat gained by the water:

Q₁ = m₁c₁ΔT₁₂

Q₁ = (5000 g)(4.184 J/g·°C)(35°C - 20°C)

Q₁ = 5000 × 4.184 × 15

Q₁ = 313,800 J

Therefore, the heat gained by the water is 313,800 J.

Next, let's calculate the heat lost by the metal:

Given:

• Mass of metal (m₂) = 2 kg
• Initial temperature of metal (T₃) = 100°C
• Final temperature of metal (T₄) = 35°C
• Specific heat capacity of metal (c₂) = 0.4 J/g·°C

Using the formula Q = mcΔT, we can calculate the heat lost by the metal:

Q₂ = m₂c₂ΔT₃₄

Q₂ = (2000 g)(0.4 J/g·°C)(35°C - 100°C)

Q₂ = 2000 × 0.4 × -65

Q₂ = -52,000 J

Therefore, the heat lost by the metal is -52,000 J (negative sign indicates heat loss).

To summarize:

• Heat gained by the water: 313,800 J
• Heat lost by the metal: -52,000 J

Note: The negative sign for the heat lost by the metal indicates that energy is being transferred out of the system (metal block), thus resulting in a decrease in temperature.