A mass of 286.8 g with unknown composition is heated in an oven to a temperature

Question

A mass of 286.8 g with unknown composition is heated in an oven to a temperature of 325.4 °C. The mass is then dropped into a strong insulated calorimeter containing 865.9 g of water. The initial temperature of water is 11.82 °C. The final equlibrium temperature of the system is 17.39 °C. Assume that the mass and water are in an isolated system and the water does not vaporize. No heat is transferred to the water during this experiment. The specific heat of the water is 4190 J/kg-K. What is the specific heat of this unknown mass?

Explanation / Answer

Mass of unknown composition, m1 = 286.8 g = 0.2868 Kg
Its intial temperature t1 = 325.4 0C Let s1 be its specific heat. Mass of water m2 = 865.9 g = 0.8659 Kg Its intial temperature t2 = 11.82 0C Specifc heat of water s2 = 4190 J/Kg K Equilibrium temperature t = 17.390C From law of calorimetry                 heat lost by unknown material = heat gained by water                                            m1 s1 (t1 - t) = m2 s2 (t - t2)         (0.2868 Kg)s1(325.4 0C -17.390C) = (0.8659 Kg)(4190 J/Kg K)(17.390C-11.82 0C)                                                 88.337 s1 = 20208.63                                                              s1 = 228.76 J/Kg K

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