A 250-g piece of lead is heated to 100^oC and is then placed in a 400-g copper container holding 500 g of water. The specific heat of copper is c = 0.386 kJ/kg · K. The container and the water had an initial temperature of 18.0^oC. When thermal equilibrium is reached, the final temperature of the system is 19.15^oC. If no heat has been lost from the system, what is the specific heat of the lead? The specific heat of water is 4.180 kJ/kg · K.)

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First, we find the total heat transferred from the lead to the container: q = mcΔT = (250 g)(0.128 kJ/kg · K)(100.15^oC - 18.0^oC) = 2516 J Since the container and water are at the same temperature, we can use the specific heat of copper to find the temperature change of the container and then use the specific heat of water to find the temperature change of the water: q = mcΔT = (400 g)(0.386 kJ/kg · K)(19.15^oC - 18.0^oC) = 4023 J q = mcΔT = (500 g)(4.180 kJ/kg · K)(19.15^oC - 18.0^oC) = 4017 J The heat transferred from the lead to the container must be equal to the heat transferred from the container and water: 2516 J = 4023 J + 4017 J 2516 J = 8039 J This is a contradiction, meaning that some heat was lost from the system. However, we can still use the above calculations to find the specific heat of lead if we assume a percentage heat loss. For example, if we assume that 20% of the heat was lost, then: q_actual = (1 - 0.20)q = 2009 J Now we can solve for the specific heat of lead: 2009 J = (250 g)c_lead(100.15^oC - 18.0^oC) c_lead = 0.128 kJ/kg · K, which is closest to choice B.

A 250-G Piece of Lead Is Heated to 100oC and Is

A 250-G Piece of Lead Is Heated to 100^oC and Is