calculate the specific heat capacity of thallium

Question

A calorimeter designed to have negligible heat loss is used to determine the specific heat capacity of

metals. A piece of thallium having a mass of 111.2 g is warmed to 95.0ºC and placed into the calorimeter

containing 125.00 g of water at 12.5ºC. The water temperature goes up to 14.9ºC. Use this information

to calculate the specific heat capacity of thallium. The specific heat capacity of water is 4.184 J/g·ºC.

Answer 1

Based on the law of conservation of energy, the heat LOST by the hot metal must be equal to the heat GAINED by the cooler water. The relationship for this is the following:

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

For thallium: q = mC∆T

q = (111.2 g)(C)(80.1º) .. the value of ∆T comes from the difference between 95º and 14.9º

For H2O: q = mC∆T

q = (125 g)(4.184 J/gº)(2.4º) .. the value of ∆T comes from the difference between 14.9º and 12.5º

Conservation of energy:

q lost by thallium = q gained by H2O

(111.2 g)(C)(80.1º) = (125 g)(4.184 J/gº)(2.4º)

8907C = 1255

C = 0.141 J/gº = specific heat of thallium