pre-lab Questions1.When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?The specific heat of water is 4.184J/g oC. Show the calculation works.Q= C x M x T3kg= 3000g4.18j/gc x 3000g x 70c = 877800 Joules= 877.8 Kj877.8Kj of energy is required to warm 3kg of water from 10C to 80C2.A calorimeter contains 100 g of water at 39.8 ºC. A 8.23 g object at 50 ºC is placed insidethe calorimeter. When equilibrium has been reached, the new temperature of the water and metal object is 40 ºC. Calculate the specific heat of the metal. The specific heat of water is 4.184J/g oC. Show the calculation works. 100g *4.18*(40-39.8)˚C = 8.23*C m*(50-40)˚C=1.02J/g Enthalpy and Specific Heat
experiment 1: determination of specific heat of a metalData SheetTable 2: MassMass (g(Water49.5g Unknown Metal Strip11.8gTable 3: Specific Heat DataTime (minutes(Temperature (°C(Trial 1Trial 2Trial 3Initial23.3 C24.1 C23.5 C5minutes23.4 C24.4 C25.0 C6minutes23.4 C24.4 C24.9 C7minutes23.5 C24.4 C24.9 C 8minutes23.5 C24.4 C24.9. C9minutes23.5 C24.4 C24.9 C 10minutes23.5 C24.4 C24.9Specific Heat Capacity of the Unknown Metal1.771.180.25Average Specific Heat Capacity of the Unknown Metal1.066Post-Lab Questions1.Calculate the specific heat capacity of the unknown metal for all three trials. The specific heat of water is 4.184J/g oC. Show the calculation works. Hint:initial temperature of the unknown metal is 100 oC.