A $170.0 \mathrm{~g}$ sample of a substance is heated to $120.0^{\circ} \mathrm{C}$ and then plunged * 4 points into $200.0 \mathrm{~mL}$ of water at $10.0^{\circ} \mathrm{C}$. The resulting mixture has a temperature of $12.6^{\circ} \mathrm{C}$. What is the specific heat capacity of the substance?

Step 1 ：Given: m_substance = 170 g, T_initial_substance = 120 °C, m_water = 200 g, T_initial_water = 10 °C, T_final_mixture = 12.6 °C, c_water = 4.18 J/g°C

Step 2 ：Using the formula q_substance = q_water, we have: 170 * c_substance * (12.6 - 120) = 200 * 4.18 * (12.6 - 10)

Step 3 ：Solving for c_substance, we get: c_substance = \(\frac{200 * 4.18 * (12.6 - 10)}{170 * (12.6 - 120)}\)

Step 4 ：Calculating the value, we get: c_substance = -0.119 J/g°C

Step 5 ：\(\boxed{-0.119}\) J/g°C is not physically meaningful. There might be an error in the given data or the problem statement.