A chemist mixes 500 g of lead at 500°C with 1,200 g of water at 20°C. She then mixes 500 g of copper at 500°C with 1,200 g of water at 20°C. The specific heat capacity of lead is 0.1276 J/g°C and the specific heat capacity of copper is 0.3845 J/g°C. What will be true about the final temperatures of the two systems? A. The final temperature of the lead-water system will be equal to the final temperature of the copper-water system. B. The final temperature of the lead-water system will be higher than the final temperature of the copper-water system. C. The final temperature of the lead-water system will be lower than the final temperature of the copper-water system. D. Cannot be determined from the given information.
Question
A chemist mixes 500 g of lead at 500°C with 1,200 g of water at 20°C. She then mixes 500 g of copper at 500°C with 1,200 g of water at 20°C. The specific heat capacity of lead is 0.1276 J/g°C and the specific heat capacity of copper is 0.3845 J/g°C. What will be true about the final temperatures of the two systems? A. The final temperature of the lead-water system will be equal to the final temperature of the copper-water system. B. The final temperature of the lead-water system will be higher than the final temperature of the copper-water system. C. The final temperature of the lead-water system will be lower than the final temperature of the copper-water system. D. Cannot be determined from the given information.
Solution
To answer this question, we need to understand that when two substances are mixed, the heat lost by the hot substance is equal to the heat gained by the cold substance. This is based on the principle of conservation of energy.
The formula for heat transfer is Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the lead-water system, the heat lost by the lead will be equal to the heat gained by the water. So, we have:
Q_lead = Q_water
m_leadc_lead(T_initial_lead - T_final) = m_waterc_water(T_final - T_initial_water)
For the copper-water system, the heat lost by the copper will be equal to the heat gained by the water. So, we have:
Q_copper = Q_water
m_copperc_copper(T_initial_copper - T_final) = m_waterc_water(T_final - T_initial_water)
Given that the initial temperatures and masses of the lead, copper, and water are the same in both systems, and the specific heat capacity of water is the same in both systems, the only difference between the two equations is the specific heat capacity of the lead and copper.
Since the specific heat capacity of copper is higher than that of lead, copper will lose more heat than lead for the same change in temperature. This means that the final temperature of the copper-water system will be lower than the final temperature of the lead-water system.
So, the correct answer is C. The final temperature of the lead-water system will be lower than the final temperature of the copper-water system.
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