A 0.056 kg sample of an unknown substance is heated to 89.6 °C in a hot water tube. It is then quickly removed and placed into an insulated jar holding 188 ml of water initially at 132.8° F. The final equilibrium temperature of the system is 67° C. What is the specific heat of this unknown substance?

The specific heat of the unknown substance is 6.84 J/gºC How do I determine the specific heat capacity?First, we shall convert 132.8° F to ° C. Thisis illustrated below:Temperature (° F) = 132.8 ° FTemperature (° C) =?°C = (°F − 32) × 5/9 °C = (132.8 − 32) × 5/9 °C = 56 °C Next, we shall determine the heat absorbed by the water. This is illustrated below:Volume of water (V) = 188 mLDensity of water (d) = 1 g/mLMass of water (M) = dV = 1 × 188 = 188 gInitial temperature (T₁) = 56 °CFinal temperature (T₂) = 67 °CChange in temperature (ΔT) = 67 - 56 = 11 °CSpecific heat capacity of copper (C) = 4.184 J/gºC Heat absorbed (Q) =?Q = MCΔT Q = 188 × 4.184 × 11Q = 8652.512 JFinally, we shall determine the specific heat capacity of the unknown metal. This can be obtained as follow:Heat absorbed (Q) = 8652.512 JHeat released (Q) = -8652.512 JMass of unknown substance (M) = 0.056 kg = 56 gInitial temperature (T₁) = 89.6 °CFinal temperature (T₂) = 67 °CChange in temperature (ΔT) = 67 - 89.6 = -22.6 °CSpecific heat capacity of unknown substance (C) = ? Q = MCΔT-8652.512 = 56 × C × -22.6-8652.512 = C × -1265.6Divide both sides by -1265.6C = -8652.512 / -1265.6C = 6.84 J/gºC Thus, the specific heat capacity is 6.84 J/gºC Learn more about specific heat capacity:https://brainly.com/question/19104255#SPJ1