Question

3. A sample of metal weighing 35.5g at a temperature of 100 oC was placed in a calorimeter containing 50 g of water at 25.0 oC. At equilibrium the temperature of water and metal was 35.5 oC. Calculate the heat capacity of the metal. . Use this value for Cp:27.65.

4. When 0.50 g of magnesium metal is placed in a calorimeter, and 100. mL of 1.0 M HCl were added the temperature of solution increased from 22.2 oC to 44.2 oC. Assume that the specific heat of solution is 4.18 J/goC and that the density is 1g/mL. Write a balanced equation for the above reaction. Calculate the enthalpy of reaction per mole of magnesium. . Use your value for Cp:27.65.

step by step please

Answer #1

3. Heat lost by the metal = Heat gained by the calorimeter

Heat lost by the metal = m*c*deltaT

m = mass of metal = 35.5g

c = specific heat capacity of metal

DeltaT = change in temperature = 100°C - 35.5°C = 64.5°C

So, heat lost by metal = 35.5*c*64.5 = 2289.75c

Heat gained by water = m*cp*deltaT

m = mass of water = 50g

cp = specific heat capacity of water = 27.65

DeltaT = 35.5°C - 25°C = 10.5°C

Heat gained by water = 50g*27.65*10.5 = 14516.25

Heat lost = heat gained

2289.75c = 14516.25

c = 14516.25 / 2289.75

= 6.34

Heat capacity of metal = 6.34J/g°C

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