please help. thanks A 250-g piece of lead is heated to 100 degree C and is then
ID: 2294768 • Letter: P
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please help. thanks
A 250-g piece of lead is heated to 100 degree C 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 degree C. When thermal equilibrium is reached, the final temperature of the system is 19.15 degree C. 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.) 0.119 kJ/kg . K 0.128 kJ/kg . K 0.110 kJ/kg . K 0.0866 kJ/kg. K 0.0372 kJ/kg . K.Explanation / Answer
15) Option B) 0.28 kJ/kg
m_lead = 0.25 kg, C_lead = ?
m_water = 0.5 kg, C_water = 4186 J/kg.K
m_copper = 0.4 kg, C_copper = 386 J/kg.K
heat lost by lead = heat gained by water and coper
m_lead*C_lead*(100-19.15) = m_water*C_water*(19.15-18) + m_copper*c_copper*(19.15-18)
0.25*80.85*C_lead = 0.5*4186*1.15 + 0.4*386*1.15
C_lead = (0.5*4186*1.15 + 0.4*386*1.15)/(0.25*80.85)
= 128 J/kg.K
= 0.128 kJ/kg.K
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