Question

A 2.74g sample of a substance assumed to be pure gold is warmed to 80.0C and submerged in to 15.2 g of water, initially at 24.7 C. The final temperature of the mixture is 25.30C. What is the heat capacity of the substance? could this substance be pure gold?

Answer 1

m(water) = 15.2 g

T(water) = 24.7 oC

C(water) = 4.184 J/goC

m(gold) = 2.74 g

T(gold) = 80.0 oC

C(gold) = to be calculated

We will be using heat conservation equation

use:

heat lost by gold = heat gained by water

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

2.74*C(gold)*(80.0-25.3) = 15.2*4.184*(25.3-24.7)

149.878*C(gold) = 38.1581

C(gold)= 0.255 J/goC

Answer: 0.255 J/goC

Actual specific heat capacity of gold is 0.129 J/goC

So, this can’t be pure gold