Question

A 48.2 g sample of a metal is heated to 95.8 degrees C and placed in a coffee-cup calorimeter containing 79.0 g of water at a temperature of 18.5 degrees C. After the metal cools, the final temperature of the metal and water is 22.8 degrees C. Calculate the specific heat capacity of the metal, assuming that no heat escapes to the surroundings or is transferred to the calorimeter.

Answer #1

Heat lost by metal = Heat gained by water

mcdt = m'c'dt'

Where

m = mass of metal = 48.2 g

c = specific heat capacity of metal = ?

dt = change in temperature of metal = initial - final

= 95.8 - 22.8

= 73 oC

m' = mass of water = 79.0 g

c' = specific heat capacity of water = 4.186 J/g-oC

dt' = change in temperature of water = 22.8-18.5 = 4.3 oC

Plug the values we get

c = ( m'c'dt') / ( mdt)

= 0.404 J/(g-oC)

Therefore the specific heat capacity of the metal is 0.404 J/(g-oC)

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