Question

A 91.70 g sample of metal was heated in a boiling water bath at 99.4 °C. The hot metal was then placed in a calorimeter, with heat capacity 39 J/K containing 45.0 g of water. Analysis of the thermogram showed the initial temperature to be 21.1°C, the final temperature to be 42.3°C.

Calculate Delta T water

Calculate Delta T metal

Calculate . CsPmetal

Calculate M (there's a - above the M.. don't know what that means) metal

What is the likely identity of the metal?

PLEASE post the material as I want to learn this material and not use chegg just to get answers.

Answer 1

Increase in the temperature of water inside calorimeter, T = Tf - Ti =  42.3°C - 21.1°C = 21.2 °C or 21.2 K

mass of water inside calorimeter, m = 45.0 g

specific heat capacity of water, S = 4.184 J/g.°C

=> Heat absorbed by water inside the calorimeter, Qw = mxSxT = 45.0 g x 4.184 J/g.°C x 21.2 °C

= 3991.54 J

Given the heat capacity of calorimeter, C = 39 J/K

increase in the temperature of calorimeter, T =  42.3°C - 21.1°C = 21.2 °C or 21.2 K

=> heat absorbed by calorimeter, Qc = C x T = 39 J/K x 21.2 K = 826.8 J

Hence total heat absorbed by calorimeter-water combination,

Qt = Qw + Qc = 3991.54 J + 826.8 J = 4818.34 J

Given the mass of the unknown metal taken, m = 91.70 g

initial temperature of the unknown metal, Ti = temperature of hot water = 99.4 °C

Final temperature of the metal, Tf = Tf of calorimeter = 42.3°C

Hence decrease in the temperature of metal, T = Tf - Ti = 42.3°C - 99.4 °C = - 57.1 °C   

Let the specific heat capacity of metal be 'S'

=> Heat released by the metal, Q = mxSxT = 91.70 g x S x (- 57.1 °C)

Also Q = - (heat absorbed by calorimeter-water combination) = - Qt

=> 91.70 g x S x (- 57.1 °C) = - 4818.34 J

=> S = 0.91 J/g.°C

The specific heat capacity of Al is also 0.91 J/g.°C.

Hence the unknown metal is aluminum (Al) (answer)