The heat required to raise 3kg of copper from 0°C to 10°C raises 1 kg of lead from 10°C to 100°C.

Question

If the specific heat capacity of copper is 0.095 cal/g°C, find the specific heat capacity of lead.

Answer 1

An elegant way to solve this is to simply equate the two conditions with each other and then solve for the specific heat of lead. We are able to accomplish this because the question tells us there is an equivalent amount of energy transferred via heat in both scenarios. In other words, the copper in the first scenario absorbs an equivalent amount of energy via heat as the lead in the second scenario:

q_Cu = q_Pb

m_CuC_CuΔT_Cu = m_PbC_PbΔT_Pb => C_Pb = m_CuC_CuΔT_Cu/m_PbΔT_Pb

<=> (3000g)(0.095 cal/g°C)(10°C - 0°C)/(1000g)(100°C - 10°C)

C_Pb = 0.03167 cal/g°C