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A 2.69 g sample of a substance suspected of being pure gold is warmed to 72.6...

A 2.69 g sample of a substance suspected of being pure gold is warmed to 72.6 ∘C and submerged into 14.4 g of water initially at 24.9 ∘C. The final temperature of the mixture is 26.8 ∘C. (Assume that the specific heat capacity of water is 4.184 Jg−1∘C−1.) What is the heat capacity of the unknown substance?

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Answer #1

m(water) = 14.4 g

T(water) = 24.9 oC

C(water) = 4.184 J/goC

m(substance) = 2.69 g

T(substance) = 72.6 oC

C(substance) = to be calculated

We will be using heat conservation equation

use:

heat lost by substance = heat gained by water

m(substance)*C(substance)*(T(substance)-T) = m(water)*C(water)*(T-T(water))

2.69*C(substance)*(72.6-26.8) = 14.4*4.184*(26.8-24.9)

123.202*C(substance) = 114.4742

C(substance)= 0.9292 J/goC

This is specific heat capacity of unknown

Heat capacity = m(substance)*C(substance)

= 2.69 g * 0.9292 J/goC

= 2.50 J/oC

Answer: 2.50 J/oC

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