91Ó°ÊÓ

Heat will be transferred from the unknown substance to the calorimeter and water until they reach thermal equilibrium. From this principle, we know that the heat gained by the water and the calorimeter is equal to the heat lost by the hot substance. Mathematically, \(Q_{substance} = Q_{water} + Q_{calorimeter}\), where Q is the heat transferred.

The heat transferred is given by \(Q = mcΔT\), where m is the mass, c is the specific heat, and ΔT is the change in temperature. The specific heats of water and aluminum (material of the calorimeter) are \(4.18 \mathrm{~kJ/kg~K}\) and \(0.897 \mathrm{~kJ/kg~K}\) respectively. \n For the unknown substance: \(ΔT_{substance} = T_{initial_{substance}} - T_{equilibrium} = 95.0 \mathrm{~C} - 32.0\mathrm{~C} = 63.0 \mathrm{~K}\). The heat is \(Q_{substance}= m_{substance}\cdot c_{substance}\cdot ΔT_{substance} = 0.125 \mathrm{~kg} \cdot c_{substance} \cdot 63.0 \mathrm{~K}\), \n For the water: \(ΔT_{water} = T_{equilibrium} -T_{initial_{water}}= 32.0\mathrm{~C} - 25\mathrm{~C}=7\mathrm{~K}\). The heat is \(Q_{water}=m_{water}\cdot c_{water}\cdot ΔT_{water} = 0.285\mathrm{~kg} \cdot 4.18\mathrm{~kJ/kg~K} \cdot 7.0\mathrm{~K}\), \n For the calorimeter, \(ΔT_{calorimeter} = ΔT_{water} = 7\mathrm{~K}\). The heat is \(Q_{calorimeter}= m_{calorimeter}\cdot c_{calorimeter}\cdot ΔT_{calorimeter} = 0.150\mathrm{~kg} \cdot 0.897\mathrm{~kJ/kg~K}\cdot 7\mathrm{~K}\).

Setting up the equation from Step 1 and substituting the equations from Step 2, we end up with \(0.125 \mathrm{~kg} \cdot c_{substance} \cdot 63 \mathrm{~K} = 0.285\mathrm{~kg} \cdot 4.18\mathrm{~kJ/kg~K} \cdot 7.0\mathrm{~K} + 0.150\mathrm{~kg} \cdot 0.897\mathrm{~kJ/kg~K}\cdot 7\mathrm{~K}\). Solving this equation for \(c_{substance}\) gives the specific heat of the unknown substance.