CHM 2045L- Heat Effects and Calorimetry Prelab: Heat Effects and Calorimetry Nam

Question

CHM 2045L- Heat Effects and Calorimetry Prelab: Heat Effects and Calorimetry Name: Class period: Show calculations and be mindful of significant figures for full credit. Date: 1. A sample ofa metal with a mass of 2.74 g that might be gold was warmed to 72.1°C It was put into a Styrofoam calorimeter where the 15.2 g of water present was initially at 24.7C. At equilibrium, the final temperature was recorded to be 26.3*C. Answer the following questions: 2.74 Mass of the metal Mass of the water Initial temperature of metal Initial temperature of water Final temperature ATwater (0.5 pt.) ATmetal (0.5 pt.) 15.2 g 72.1°C 24.7 C 26.3? quzo (S.H. water-4.18 J/g C) use equation 1 (4 pt.) qmetal use equation 2 (0.5 pt.) Specific heat of metal use equation 3 (4 pt.) J/g C Is it gold? (see text table 6.4) (O.5 pt. of gold Is this gold? Yes No (circle answer) (Continued next page)

Explanation / Answer

1. specific heat of metal

mass of metal = 2.74 g

mass of H2O = 15.2 g

change in temperature dT(H2O) = 26.3 - 24.7 = 1.6 oC

change in temperature dT(meta;) = 72.1 - 26.3 = 45.8 oC

q(H2O) = mCpdT = 15.2 x 4.18 x 1.6 = 101.6576 J

q(metal) = mCpdT = 2.74 x Cp x 45.8 = 101.6576 J

specific heat of metal (Cp) = 0.810 J/g.oC

Literature specific heat of gold = 0.129 J/g.oC

So the metal under study is not gold.

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2. heat of solution

The reaction in solution = exothermic

mass H2O = 53 g

mass solid = 2 g

initial temperature = 24 oC

final temperature = 34.7 oC

dT = 34.7 - 24 = 10.7 oC

q(soln) = mCpdT = 55 x 4.18 x 10.7 = 2460 J

dH(soln) = -2460 J

dH(soln) per gram solid = -2460 J/2 g x 1000 = 1.23 kJ/g

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