Question

A 2.44 g sample of a substance suspected of being pure gold is warmed to 72.5 ∘C and submerged into 15.0 g of water initially at 24.8 ∘C. The final temperature of the mixture is 26.4 ∘C.

What is the heat capacity of the unknown substance?Express your answer using two significant figures.

Could the substance be pure gold?

Answer 1

1)

m(water) = 15.0 g

T(water) = 24.8 oC

C(water) = 4.184 J/goC

m(sample) = 2.44 g

T(sample) = 72.5 oC

C(sample) = to be calculated

We will be using heat conservation equation

use:

heat lost by sample = heat gained by water

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

2.44*C(sample)*(72.5-26.4) = 15.0*4.184*(26.4-24.8)

112.484*C(sample) = 100.416

C(sample)= 0.8927 J/goC

Answer: 0.89 J/goC

2)

Specific heat capacity of pure gold is 0.129 J/g.oC which is much less than calculated value

So, substance can’t be pure gold

Answer: NO