Question

A 52.9 g sample of iron is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The iron sample starts off at 98.3 °C and the temperature of the water starts off at 24.0 °C. When the temperature of the water stops changing it's 26.5 °C. The pressure remains constant at 1 atm. insulated container water Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to 2 significant digits. sample a calorimeter • °C Scientific Notation

Answer 1

We can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙^oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (^oC Celcius)

Heat lost from Iron = Heat gained by water

For Iron

m = 52.9 gm c = ?, Ti =98.3 ^oC Tf = 26.5 ^oC

∆T = 98.3 ^oC - 26.5 ^oC = 71.8 ^oC

For water

m = 200 gm c = 4.184 J/g∙^oC, Ti = 24 ^oC Tf = 26.5 ^oC

∆T = 26.5 ^oC - 24 ^oC = 2.5 ^oC

52.9 gm x Sp Heat x 71.8 ^oC = 200 gm  x 4.184 J/g∙^oC x 2.5 ^oC

Sp Heat = 0.5507 J/g∙^oC

The specific heat capacity of the iron is 0.5507 J/g∙^oC